Compute $sumlimits_j = 0^m - 1 left(c_j + 1right)lnleft(c_j + 1right)$ where $c_j = cosleft(fracpi2mleft(1 + 2jright) right)$ The Next CEO of Stack OverflowAn arctan series with a parameter $sum_n=1^infty arctan left(frac2a^2n^2right)$Evaluating $int_0^1 fracx arctan x log left( 1-x^2right)1+x^2dx$Improper Integral $int_0^1left(leftfrac1xright-frac12right)fraclog(x)xdx$Derive $log(a+b)=log(a)-2logleft(cosleft(arctanleft(sqrtfracbaright)right)right)$Is there any simple method to evaluate this integral $int_0^2mpi} frac1{sin^4(x)+cos^4(x)d x?$How do I evaluate $lim_n rightarrow inftysum_k=1^nlogleft ( 1+frackn^2 right )$?On the summation $sum limits_n=1^infty arctan left ( frac1n^3+n^2+n+1 right )$Absolute convergence of $sumlimits_nleft( log (1-fraczn)^n^k + sumlimits_m=1^k+1log e^n^k-mz^m/m right)$?Solving a set of equations containing sums of logarithmsMethods to solve $int_0^infty fraccosleft(kx^nright)x^n + a:dx$
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Compute $sumlimits_j = 0^m - 1 left(c_j + 1right)lnleft(c_j + 1right)$ where $c_j = cosleft(fracpi2mleft(1 + 2jright) right)$
The Next CEO of Stack OverflowAn arctan series with a parameter $sum_n=1^infty arctan left(frac2a^2n^2right)$Evaluating $int_0^1 fracx arctan x log left( 1-x^2right)1+x^2dx$Improper Integral $int_0^1left(leftfrac1xright-frac12right)fraclog(x)xdx$Derive $log(a+b)=log(a)-2logleft(cosleft(arctanleft(sqrtfracbaright)right)right)$Is there any simple method to evaluate this integral $int_0^2mpi frac1sin^4(x)+cos^4(x)d x?$How do I evaluate $lim_n rightarrow inftysum_k=1^nlogleft ( 1+frackn^2 right )$?On the summation $sum limits_n=1^infty arctan left ( frac1n^3+n^2+n+1 right )$Absolute convergence of $sumlimits_nleft( log (1-fraczn)^n^k + sumlimits_m=1^k+1log e^n^k-mz^m/m right)$?Solving a set of equations containing sums of logarithmsMethods to solve $int_0^infty fraccosleft(kx^nright)x^n + a:dx$
$begingroup$
As part of solving:
beginequation
I_m = int_0^1 lnleft(1 + x^2mright):dx.
endequation
where $m in mathbbN$. I found an unresolved component that I'm unsure how to start:
beginequation
G_m = sum_j = 0^m - 1 left(c_j + 1right)lnleft(c_j + 1right),
endequation
where $c_j = cosleft(fracpi2mleft(1 + 2jright) right)$
I'm just looking for a starting point. Any tips would be greatly appreciated.
By the way, I was able to show (and this was part of the solution too) :
beginequation
sum_j = 0^m - 1 c_j = 0
endequation
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
beginequation
int_0^1 frac1t^n + 1:dt = frac1nleft[fracpisinleft(fracpin right)- Bleft(1 - frac1n, frac1n, frac12right)right]
endequation
Or for any positive upper bound $x$:
beginalign
I_n(x) &= int_0^x frac1t^n + 1:dt = frac1nleft[Gammaleft(1 - frac1n right)Gammaleft(frac1n right)- Bleft(1 - frac1n, frac1n, frac1x^n + 1right)right]
endalign
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m in mathbbN$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
beginalign
x^2m + 1 = 0 rightarrow x^2m = e^pi i
endalign
By De Moivre's formula, we observe that:
beginalign
x = expleft(fracpi + 2pi j2m i right) mbox for j = 0dots 2m - 1,
endalign
which we can express as the set
beginalign
S &= Bigg expleft(fracpi + 2pi cdot 02m i right) , :expleft(fracpi + 2pi cdot 12m i right),dots,:expleft(fracpi + 2pi cdot (2m - 2)2m i right)\
&qquad:expleft(fracpi + 2pi cdot (2m - 1)2m i right)Bigg,
endalign
which can be expressed as the set of $2$-tuples
beginalign
S &= left left( expleft(fracpi + 2pi j2m i right) , :expleft(fracpi + 2pi(2m - 1 - j )2m i right)right): bigg\
& = left (z_j, cleft(z_jright):
endalign
From here, we can factor $x^2m + 1$ into the form
beginalign
x^2m + 1 &= prod_r in S left(x + r_jright)left(x + c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
endalign
For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(fracpi + 2pi j2m right)= cosleft(fracpi2mleft(1 + 2jright)right) = c_j$
beginalign
int_0^1 logleft( x^2m + 1right):dx &= int_0^1 logleft(prod_r in S left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_j = 0^m - 1 int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_j = 0^m - 1 left[2sqrt1 - c_j^2arctanleft(fracx + c_jsqrt1 - c_j^2right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_j = 0^m - 1 left[ 2sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_j = 0^m - 1sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
Thus,
beginalign
int_0^1 logleft( x^2m + 1right):dx &=sum_j = 0^m - 1c_jsinleft(fracpi2mleft(1 + 2jright)right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
integration sequences-and-series definite-integrals logarithms trigonometric-series
edited Dec 29 '18 at 11:02
Did
249k23226466
$endgroup$
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show 7 more comments
$begingroup$
As part of solving:
beginequation
I_m = int_0^1 lnleft(1 + x^2mright):dx.
endequation
where $m in mathbbN$. I found an unresolved component that I'm unsure how to start:
beginequation
G_m = sum_j = 0^m - 1 left(c_j + 1right)lnleft(c_j + 1right),
endequation
where $c_j = cosleft(fracpi2mleft(1 + 2jright) right)$
I'm just looking for a starting point. Any tips would be greatly appreciated.
By the way, I was able to show (and this was part of the solution too) :
beginequation
sum_j = 0^m - 1 c_j = 0
endequation
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
beginequation
int_0^1 frac1t^n + 1:dt = frac1nleft[fracpisinleft(fracpin right)- Bleft(1 - frac1n, frac1n, frac12right)right]
endequation
Or for any positive upper bound $x$:
beginalign
I_n(x) &= int_0^x frac1t^n + 1:dt = frac1nleft[Gammaleft(1 - frac1n right)Gammaleft(frac1n right)- Bleft(1 - frac1n, frac1n, frac1x^n + 1right)right]
endalign
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m in mathbbN$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
beginalign
x^2m + 1 = 0 rightarrow x^2m = e^pi i
endalign
By De Moivre's formula, we observe that:
beginalign
x = expleft(fracpi + 2pi j2m i right) mbox for j = 0dots 2m - 1,
endalign
which we can express as the set
beginalign
S &= Bigg expleft(fracpi + 2pi cdot 02m i right) , :expleft(fracpi + 2pi cdot 12m i right),dots,:expleft(fracpi + 2pi cdot (2m - 2)2m i right)\
&qquad:expleft(fracpi + 2pi cdot (2m - 1)2m i right)Bigg,
endalign
which can be expressed as the set of $2$-tuples
beginalign
S &= left left( expleft(fracpi + 2pi j2m i right) , :expleft(fracpi + 2pi(2m - 1 - j )2m i right)right): bigg\
& = left (z_j, cleft(z_jright):
endalign
From here, we can factor $x^2m + 1$ into the form
beginalign
x^2m + 1 &= prod_r in S left(x + r_jright)left(x + c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
endalign
For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(fracpi + 2pi j2m right)= cosleft(fracpi2mleft(1 + 2jright)right) = c_j$
beginalign
int_0^1 logleft( x^2m + 1right):dx &= int_0^1 logleft(prod_r in S left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_j = 0^m - 1 int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_j = 0^m - 1 left[2sqrt1 - c_j^2arctanleft(fracx + c_jsqrt1 - c_j^2right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_j = 0^m - 1 left[ 2sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_j = 0^m - 1sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
Thus,
beginalign
int_0^1 logleft( x^2m + 1right):dx &=sum_j = 0^m - 1c_jsinleft(fracpi2mleft(1 + 2jright)right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
integration sequences-and-series definite-integrals logarithms trigonometric-series
edited Dec 29 '18 at 11:02
Did
249k23226466
$endgroup$
This question had a bounty worth +500
reputation from Community♦ that ended ended at 2019-03-29 11:41:44Z">18 hours ago. Grace period ends in 5 hours
One or more of the answers is exemplary and worthy of an additional bounty.
$begingroup$
@Mason - Yes, thanks for the pickup, I will edit now
$endgroup$
– user150203
Dec 27 '18 at 4:46
$begingroup$
And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
$endgroup$
– Mason
Dec 27 '18 at 4:58
$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
$endgroup$
– user150203
Dec 27 '18 at 5:02
1
$begingroup$
Thanks for the shout-out David ;) I'll try to answer the question
$endgroup$
– clathratus
Dec 27 '18 at 9:08
1
$begingroup$
@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
$endgroup$
– user150203
Dec 27 '18 at 9:13
|
show 7 more comments
$begingroup$
As part of solving:
beginequation
I_m = int_0^1 lnleft(1 + x^2mright):dx.
endequation
where $m in mathbbN$. I found an unresolved component that I'm unsure how to start:
beginequation
G_m = sum_j = 0^m - 1 left(c_j + 1right)lnleft(c_j + 1right),
endequation
where $c_j = cosleft(fracpi2mleft(1 + 2jright) right)$
I'm just looking for a starting point. Any tips would be greatly appreciated.
By the way, I was able to show (and this was part of the solution too) :
beginequation
sum_j = 0^m - 1 c_j = 0
endequation
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
beginequation
int_0^1 frac1t^n + 1:dt = frac1nleft[fracpisinleft(fracpin right)- Bleft(1 - frac1n, frac1n, frac12right)right]
endequation
Or for any positive upper bound $x$:
beginalign
I_n(x) &= int_0^x frac1t^n + 1:dt = frac1nleft[Gammaleft(1 - frac1n right)Gammaleft(frac1n right)- Bleft(1 - frac1n, frac1n, frac1x^n + 1right)right]
endalign
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m in mathbbN$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
beginalign
x^2m + 1 = 0 rightarrow x^2m = e^pi i
endalign
By De Moivre's formula, we observe that:
beginalign
x = expleft(fracpi + 2pi j2m i right) mbox for j = 0dots 2m - 1,
endalign
which we can express as the set
beginalign
S &= Bigg expleft(fracpi + 2pi cdot 02m i right) , :expleft(fracpi + 2pi cdot 12m i right),dots,:expleft(fracpi + 2pi cdot (2m - 2)2m i right)\
&qquad:expleft(fracpi + 2pi cdot (2m - 1)2m i right)Bigg,
endalign
which can be expressed as the set of $2$-tuples
beginalign
S &= left left( expleft(fracpi + 2pi j2m i right) , :expleft(fracpi + 2pi(2m - 1 - j )2m i right)right): bigg\
& = left (z_j, cleft(z_jright):
endalign
From here, we can factor $x^2m + 1$ into the form
beginalign
x^2m + 1 &= prod_r in S left(x + r_jright)left(x + c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
endalign
For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(fracpi + 2pi j2m right)= cosleft(fracpi2mleft(1 + 2jright)right) = c_j$
beginalign
int_0^1 logleft( x^2m + 1right):dx &= int_0^1 logleft(prod_r in S left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_j = 0^m - 1 int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_j = 0^m - 1 left[2sqrt1 - c_j^2arctanleft(fracx + c_jsqrt1 - c_j^2right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_j = 0^m - 1 left[ 2sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_j = 0^m - 1sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
Thus,
beginalign
int_0^1 logleft( x^2m + 1right):dx &=sum_j = 0^m - 1c_jsinleft(fracpi2mleft(1 + 2jright)right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
integration sequences-and-series definite-integrals logarithms trigonometric-series
edited Dec 29 '18 at 11:02
Did
249k23226466
$endgroup$
As part of solving:
beginequation
I_m = int_0^1 lnleft(1 + x^2mright):dx.
endequation
where $m in mathbbN$. I found an unresolved component that I'm unsure how to start:
beginequation
G_m = sum_j = 0^m - 1 left(c_j + 1right)lnleft(c_j + 1right),
endequation
where $c_j = cosleft(fracpi2mleft(1 + 2jright) right)$
I'm just looking for a starting point. Any tips would be greatly appreciated.
By the way, I was able to show (and this was part of the solution too) :
beginequation
sum_j = 0^m - 1 c_j = 0
endequation
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
beginequation
int_0^1 frac1t^n + 1:dt = frac1nleft[fracpisinleft(fracpin right)- Bleft(1 - frac1n, frac1n, frac12right)right]
endequation
Or for any positive upper bound $x$:
beginalign
I_n(x) &= int_0^x frac1t^n + 1:dt = frac1nleft[Gammaleft(1 - frac1n right)Gammaleft(frac1n right)- Bleft(1 - frac1n, frac1n, frac1x^n + 1right)right]
endalign
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m in mathbbN$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
beginalign
x^2m + 1 = 0 rightarrow x^2m = e^pi i
endalign
By De Moivre's formula, we observe that:
beginalign
x = expleft(fracpi + 2pi j2m i right) mbox for j = 0dots 2m - 1,
endalign
which we can express as the set
beginalign
S &= Bigg expleft(fracpi + 2pi cdot 02m i right) , :expleft(fracpi + 2pi cdot 12m i right),dots,:expleft(fracpi + 2pi cdot (2m - 2)2m i right)\
&qquad:expleft(fracpi + 2pi cdot (2m - 1)2m i right)Bigg,
endalign
which can be expressed as the set of $2$-tuples
beginalign
S &= left left( expleft(fracpi + 2pi j2m i right) , :expleft(fracpi + 2pi(2m - 1 - j )2m i right)right): bigg\
& = left (z_j, cleft(z_jright):
endalign
From here, we can factor $x^2m + 1$ into the form
beginalign
x^2m + 1 &= prod_r in S left(x + r_jright)left(x + c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_i = 0^m - 1 left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
endalign
For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(fracpi + 2pi j2m right)= cosleft(fracpi2mleft(1 + 2jright)right) = c_j$
beginalign
int_0^1 logleft( x^2m + 1right):dx &= int_0^1 logleft(prod_r in S left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_j = 0^m - 1 int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_j = 0^m - 1 left[2sqrt1 - c_j^2arctanleft(fracx + c_jsqrt1 - c_j^2right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_j = 0^m - 1 left[ 2sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_j = 0^m - 1sqrt1 - c_j^2arctanleft(sqrtfrac1 - c_j1 + c_j right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
Thus,
beginalign
int_0^1 logleft( x^2m + 1right):dx &=sum_j = 0^m - 1c_jsinleft(fracpi2mleft(1 + 2jright)right) + log(2)sum_j = 0^m - 1 c_j + mleft(log(2) - 2right)\
&qquad+ sum_j = 0^m - 1left(c_j + 1right)logleft(c_j + 1right)
endalign
integration sequences-and-series definite-integrals logarithms trigonometric-series
integration sequences-and-series definite-integrals logarithms trigonometric-series
edited Dec 29 '18 at 11:02
Did
249k23226466
edited Dec 29 '18 at 11:02
Did
249k23226466
edited Dec 29 '18 at 11:02
Did
249k23226466
edited Dec 29 '18 at 11:02
Did
249k23226466
edited Dec 29 '18 at 11:02
Did
249k23226466
249k23226466
asked Dec 27 '18 at 4:44
user150203
This question had a bounty worth +500
reputation from Community♦ that ended ended at 2019-03-29 11:41:44Z">18 hours ago. Grace period ends in 5 hours
One or more of the answers is exemplary and worthy of an additional bounty.
This question had a bounty worth +500
reputation from Community♦ that ended ended at 2019-03-29 11:41:44Z">18 hours ago. Grace period ends in 5 hours
One or more of the answers is exemplary and worthy of an additional bounty.
$begingroup$
@Mason - Yes, thanks for the pickup, I will edit now
$endgroup$
– user150203
Dec 27 '18 at 4:46
$begingroup$
And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
$endgroup$
– Mason
Dec 27 '18 at 4:58
$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
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– user150203
Dec 27 '18 at 5:02
1
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Thanks for the shout-out David ;) I'll try to answer the question
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– clathratus
Dec 27 '18 at 9:08
1
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@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
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– user150203
Dec 27 '18 at 9:13
|
show 7 more comments
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@Mason - Yes, thanks for the pickup, I will edit now
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– user150203
Dec 27 '18 at 4:46
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And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
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– Mason
Dec 27 '18 at 4:58
$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
$endgroup$
– user150203
Dec 27 '18 at 5:02
1
$begingroup$
Thanks for the shout-out David ;) I'll try to answer the question
$endgroup$
– clathratus
Dec 27 '18 at 9:08
1
$begingroup$
@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
$endgroup$
– user150203
Dec 27 '18 at 9:13
$begingroup$
@Mason - Yes, thanks for the pickup, I will edit now
$endgroup$
– user150203
Dec 27 '18 at 4:46
$begingroup$
@Mason - Yes, thanks for the pickup, I will edit now
$endgroup$
– user150203
Dec 27 '18 at 4:46
$begingroup$
And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
$endgroup$
– Mason
Dec 27 '18 at 4:58
$begingroup$
And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
$endgroup$
– Mason
Dec 27 '18 at 4:58
$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
$endgroup$
– user150203
Dec 27 '18 at 5:02
$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
$endgroup$
– user150203
Dec 27 '18 at 5:02
1
1
$begingroup$
Thanks for the shout-out David ;) I'll try to answer the question
$endgroup$
– clathratus
Dec 27 '18 at 9:08
$begingroup$
Thanks for the shout-out David ;) I'll try to answer the question
$endgroup$
– clathratus
Dec 27 '18 at 9:08
1
1
$begingroup$
@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
$endgroup$
– user150203
Dec 27 '18 at 9:13
$begingroup$
@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
$endgroup$
– user150203
Dec 27 '18 at 9:13
|
show 7 more comments
4 Answers
4
active
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votes
$begingroup$
This does not answer the question as asked in the post.
Consider
$$J_m=int log(1+x^2m),dx$$ One integration by parts gives
$$J_m=x log left(1+x^2 mright)-2mint frac x^2 m+1-1x^2 m+1,dx=x log left(1+x^2 mright)-2mx+2mint fracdxx^2 m+1$$ and
$$int fracdxx^2 m+1=x , _2F_1left(1,frac12 m;1+frac12 m;-x^2 mright)$$ where appears the Gaussian or ordinary hypergeometric function.
So
$$K_m=int_0^a log(1+x^2m),dx=a log left(1+a^2 mright)-2ma+2ma , _2F_1left(1,frac12 m;1+frac12 m;-a^2 mright)$$ and, if $a=1$,
$$I_m=int_0^1 log(1+x^2m),dx= log left(2right)-2m+2m , _2F_1left(1,frac12 m;1+frac12 m;-1right)$$ which can write
$$I_m=log (2)-Phi left(-1,1,1+frac12 mright)$$
where appears the Lerch transcendent function.
Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac12 mright)$ before any simplification
$$f(1)=fracpi 2-2$$
$$f(2)=frac14 left(pi tan left(fracpi 8right)+pi cot left(fracpi
8right)-4 sqrt2 log left(sin left(fracpi 8right)right)+4
sqrt2 log left(cos left(fracpi 8right)right)right)-4$$
$$f(3)=frac2 left(pi -sqrt3 log left(sqrt3-1right)+sqrt3 log
left(1+sqrt3right)right)left(sqrt3-1right)
left(1+sqrt3right)-6$$
$$f(4)=frac14 left(pi tan left(fracpi 16right)+pi cot left(fracpi
16right)-8 sin left(fracpi 8right) log left(sin left(frac3
pi 16right)right)+8 cos left(fracpi 8right) log left(cos
left(fracpi 16right)right)-8 cos left(fracpi 8right) log
left(sin left(fracpi 16right)right)+8 sin left(fracpi
8right) log left(cos left(frac3 pi 16right)right)right)-8$$
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
$endgroup$
– user150203
Dec 27 '18 at 8:49
$begingroup$
@DavidG. Your post is very interesting. I need to spend more time with it !
$endgroup$
– Claude Leibovici
Dec 27 '18 at 8:54
$begingroup$
Please do! I look forward to any comments you may have. Thanks again for your post.
$endgroup$
– user150203
Dec 27 '18 at 9:48
$begingroup$
Mathematica calculates the same result
$endgroup$
– stocha
Mar 22 at 19:11
add a comment |
$begingroup$
$colorbrowntextbfPreliminary notes.$
Calculation of the required sum looks more complex task than of the
issue integral.
Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.
Denote
$$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
Easy to see that the issue integral is $I_2m.$
$colorbrowntextbfClosed form of the integrals.$
The first step is the integration by parts:
$$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfracx^n-11+x^nmathrm dx,$$
$$I_n = log2 - n + nintlimits_0^1dfracmathrm dx1+x^n.tag2$$
(see also Claude Leibovici).
Then, the substitution
$$x=e^-ttag3$$
gives
$$J_n=intlimits_0^1dfracmathrm dx1+x^n=intlimits_0^inftydfrace^-t,mathrm dt1+e^-nt = sumlimits_k=0^infty(-1)^kintlimits_0^infty e^-(kn+1)t,mathrm dt=sumlimits_k=0^inftydfrac(-1)^kkn+1 ,$$
$$J_n = intlimits_0^1dfracmathrm dx1+x^n= dfrac12nleft(psi_0left(dfracn+12nright) - psi_0left(dfrac12nright)right),tag4$$
where $psi_0(x)$ is the polygamma function.
Using $(2),(4),$ easy to get the expression for the OP integral in the form of
$$boxedI_2m = log2 - 2m +dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right).tag5$$
$colorbrowntextbfTesting of the solution.$
Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.
Immediate calculations allow to check obtained expressions in the simple cases
beginalign
&J_1 = intlimits_0^1dfracmathrm dx1+x = log 2,\[4pt]
&J_2 = intlimits_0^1dfracmathrm dx1+x^2 = arctan 1 = dfracpi4,\[4pt]
&J_3 = intlimits_0^1dfracmathrm dx1+x^3
= dfrac16intlimits_0^1left(dfrac21+x - dfrac2x-11-x+x^2 + dfrac31-x+x^2right),mathrm dx\
&= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1sqrt3arctandfrac2x-1sqrt3right)bigg|_0^1
= dfracpi3sqrt3 + frac13 log2.\[4pt]
endalign
More hard cases are
beginalign
&J_4 = intlimits_0^1dfracmathrm dx1+x^4
= dfracsqrt28left(logdfracx^2 + sqrt2 x + 1x^2 - sqrt2 x + 1
- 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
&= dfracsqrt28left(logdfrac(x^2 + sqrt2 x + 1)^21+x^4 + 2arctandfracxsqrt21-x^2right)bigg|_0^1
= dfracsqrt28(pi+log(3+2sqrt2)),\[4pt]
&J_4=dfrac18Bigg(dfrac12sqrtdfrac2-sqrt22+sqrt2pi+dfrac12sqrtdfrac2+sqrt22-sqrt2pi
-2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
&+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
&=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
+sqrt2logdfrac2+sqrt22-sqrt2Bigg)
= dfracsqrt28(pi+log(3+2sqrt2)),
endalign
and $n>4.$
Numeric calculations of the issue integral $I_2m$ and its closed form $(5)$ also confirm the correctness of the closed form.
$colorbrowntextbfThe alternative approach.$
The alternative approach is considered in OP. Let us repeat it with some differences.
Polynomial factorization can be presented in the form of
beginalign
&1+x^2m = prodlimits_j=0^2m-1large left(x-e^frac2j+12mpi iright)
= prodlimits_j=0^m-1left(x^2-2xcosfrac2j+12mpi+1right),
endalign
so
beginalign
&I_2m = intlimits_0^1 ln(1+x^2m),mathrm dx
= sumlimits_j=0^m-1 T_j,quadtextwhere\[4pt]
&T_j = intlimits_0^1 lnleft(x^2-2xcosfrac2j+12mpi+1right),mathrm dx,\[4pt]
&T_j = xlnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1
- 2intlimits_0^1 dfracxleft(x-cosfrac2j+12mpiright)x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright)
- 2-2intlimits_0^1 dfracxcosfrac2j+12mpi-1x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- 2cosfrac2j+12mpi intlimits_0^1 dfracx-cosfrac2j+12mpix^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& + 2intlimits_0^1 dfracsin^2frac2j+12mpileft(x-cosfrac2j+12mpiright)^2+sin^2frac2j+12mpi,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1\[4pt]
& + 2sinfrac2j+12mpi arctandfracx-cosfrac2j+12mpisinfrac2j+12mpibigg|_0^1\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
left(arctandfrac1-cosfrac2j+12mpisinfrac2j+12mpi
+ arctandfraccosfrac2j+12mpisinfrac2j+12mpiright)\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
arctandfracsinfrac2j+12mpi
sin^2frac2j+12mpi - left(1 - cosfrac2j+12mpiright)cosfrac2j+12mpi\[4pt]
& = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + 2sinfrac2j+12mpiarctancotfrac2j+14mpi,\[4pt]
&T_j = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + dfrac2(m-j)-12mpisinfrac2j+12mpi.
endalign
Taking in account that
beginalign
&sumlimits_j=0^m-1cosfrac2j+12mpi
= Re sumlimits_j=0^m-1e^frac2j+12mpi i
= Re large dfrac1-e^pi i1-e^fracpimie^fracpi2mi
=Redfrac isinfracpi2m = 0,\[4pt]
&sumlimits_j=0^m-1sinfrac2j+12mpi
= Im sumlimits_j=0^m-1e^frac2j+12mpi i
=Imdfracisinfracpi2m = dfrac1sinfracpi2m,\[4pt]
&sumlimits_j=0^m-1frac2j+12mpisinfrac2j+12mpi
= dfracpi2sinfracpi2m
endalign
(see also Wolfram Alpha calculations),
one can get
$$boxedI_2m = m(ln2-2) + dfracpi2cscfracpi2m
+ sumlimits_j=0^m-1left(1pmcosfrac2j+12mpiright)
logleft(1pmcosfrac2j+12mpiright).tag6$$
$colorbrowntextbfClosed form for the sum.$
From $(5)-(6)$ should
$$colorgreenboxedsumlimits_j=0^m-1left(1pm c_jright)
logleft(1pm c_jright)
= dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right) - (m-1)ln2 - dfracpi2cscfracpi2m
,tag7$$
where
$$c_j = cosfrac2j+12mpi.$$
Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".
Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
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1
$begingroup$
Very nice presentation, already before the alternate approach (+1).
$endgroup$
– Markus Scheuer
yesterday
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@MarkusScheuer Thanks, you are welcome!
$endgroup$
– Yuri Negometyanov
yesterday
add a comment |
$begingroup$
I did it!
I actually have no idea whether or not this works, but this is how I did it.
$ninBbb N$
Define the sequence $r_k^(n)_k=1^k=n$ such that
$$x^n+1=prod_k=1^nbig(x-r^(n)_kbig)$$
We then know that $$r_k^(n)=expbigg[fracipin(2k-1)bigg]$$
Then we define
$$S_n=r_k^(n):kin[1,n]capBbb N$$
So we have that
$$frac1x^n+1=prod_rin S_nfrac1x-r=prod_k=1^nfrac1x-r_k^(n)$$
Then we assume that we can write
$$prod_rin S_nfrac1x-r=sum_rin S_nfracb(r)x-r$$
Multiplying both sides by $prod_ain S_n(x-a)$,
$$1=sum_rin S_nb(r)prod_ain S_n\ aneq r(x-a)$$
So for any $omegain S_n$,
$$1=b(omega)prod_ain S_n\ aneq omega(omega-a)$$
$$b(omega)=prod_ain S_n\ aneq omegafrac1omega-a$$
$$b(r_k^(n))=prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we know that
$$I_n=int_0^1fracmathrmdx1+x^n=sum_k=1^nb(r_k^(n))int_0^1fracmathrmdxx-r_k^(n)$$
$$I_n=sum_k=1^nb(r_k^(n))logbigg|fracr_k^(n)-1r_k^(n)bigg|$$
$$I_n=sum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we have
$$int_0^1log(1+x^n)mathrmdx=log2-n+nsum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
along with a plethora of other identities...
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
$endgroup$
1
$begingroup$
Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
$endgroup$
– user150203
Dec 31 '18 at 6:16
$begingroup$
I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
$endgroup$
– user150203
Dec 31 '18 at 9:47
$begingroup$
Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
$endgroup$
– clathratus
Dec 31 '18 at 20:28
add a comment |
$begingroup$
Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^(n)$ as last time, we apply the $logprod_ia_i=sum_ilog a_i$ property to see that
$$log(1+x^n)=logprod_k=1^n(x-r_k^(n))=sum_k=1^nlog(x-r_k^(n))$$
So
$$I_n=int_0^1log(1+x^n)mathrm dx=sum_k=1^nint_0^1log(x-r_k^(n))mathrm dx$$
This last integral boils down to
$$beginalign
int_0^1log(x-a)mathrm dx=&alogfraca1+a+log(1-a)-1\
=&logfraca^a(1-a)e(1+a)^a
endalign$$
So
$$I_n=sum_rin S_nlogfracr^r(1-r)e(1+r)^r$$
And you know how I love product representations, so we again use $logprod_ia_i=sum_ilog a_i$ to see that
$$
I_n=logprod_rin S_nfracr^r(1-r)e(1+r)^r\
prod_rin S_nfracr^r(1-r)(1+r)^r=exp(n+I_n)
$$
Which I just think is really neat.
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
$endgroup$
add a comment |
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4 Answers
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4 Answers
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$begingroup$
This does not answer the question as asked in the post.
Consider
$$J_m=int log(1+x^2m),dx$$ One integration by parts gives
$$J_m=x log left(1+x^2 mright)-2mint frac x^2 m+1-1x^2 m+1,dx=x log left(1+x^2 mright)-2mx+2mint fracdxx^2 m+1$$ and
$$int fracdxx^2 m+1=x , _2F_1left(1,frac12 m;1+frac12 m;-x^2 mright)$$ where appears the Gaussian or ordinary hypergeometric function.
So
$$K_m=int_0^a log(1+x^2m),dx=a log left(1+a^2 mright)-2ma+2ma , _2F_1left(1,frac12 m;1+frac12 m;-a^2 mright)$$ and, if $a=1$,
$$I_m=int_0^1 log(1+x^2m),dx= log left(2right)-2m+2m , _2F_1left(1,frac12 m;1+frac12 m;-1right)$$ which can write
$$I_m=log (2)-Phi left(-1,1,1+frac12 mright)$$
where appears the Lerch transcendent function.
Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac12 mright)$ before any simplification
$$f(1)=fracpi 2-2$$
$$f(2)=frac14 left(pi tan left(fracpi 8right)+pi cot left(fracpi
8right)-4 sqrt2 log left(sin left(fracpi 8right)right)+4
sqrt2 log left(cos left(fracpi 8right)right)right)-4$$
$$f(3)=frac2 left(pi -sqrt3 log left(sqrt3-1right)+sqrt3 log
left(1+sqrt3right)right)left(sqrt3-1right)
left(1+sqrt3right)-6$$
$$f(4)=frac14 left(pi tan left(fracpi 16right)+pi cot left(fracpi
16right)-8 sin left(fracpi 8right) log left(sin left(frac3
pi 16right)right)+8 cos left(fracpi 8right) log left(cos
left(fracpi 16right)right)-8 cos left(fracpi 8right) log
left(sin left(fracpi 16right)right)+8 sin left(fracpi
8right) log left(cos left(frac3 pi 16right)right)right)-8$$
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
$endgroup$
– user150203
Dec 27 '18 at 8:49
$begingroup$
@DavidG. Your post is very interesting. I need to spend more time with it !
$endgroup$
– Claude Leibovici
Dec 27 '18 at 8:54
$begingroup$
Please do! I look forward to any comments you may have. Thanks again for your post.
$endgroup$
– user150203
Dec 27 '18 at 9:48
$begingroup$
Mathematica calculates the same result
$endgroup$
– stocha
Mar 22 at 19:11
add a comment |
$begingroup$
This does not answer the question as asked in the post.
Consider
$$J_m=int log(1+x^2m),dx$$ One integration by parts gives
$$J_m=x log left(1+x^2 mright)-2mint frac x^2 m+1-1x^2 m+1,dx=x log left(1+x^2 mright)-2mx+2mint fracdxx^2 m+1$$ and
$$int fracdxx^2 m+1=x , _2F_1left(1,frac12 m;1+frac12 m;-x^2 mright)$$ where appears the Gaussian or ordinary hypergeometric function.
So
$$K_m=int_0^a log(1+x^2m),dx=a log left(1+a^2 mright)-2ma+2ma , _2F_1left(1,frac12 m;1+frac12 m;-a^2 mright)$$ and, if $a=1$,
$$I_m=int_0^1 log(1+x^2m),dx= log left(2right)-2m+2m , _2F_1left(1,frac12 m;1+frac12 m;-1right)$$ which can write
$$I_m=log (2)-Phi left(-1,1,1+frac12 mright)$$
where appears the Lerch transcendent function.
Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac12 mright)$ before any simplification
$$f(1)=fracpi 2-2$$
$$f(2)=frac14 left(pi tan left(fracpi 8right)+pi cot left(fracpi
8right)-4 sqrt2 log left(sin left(fracpi 8right)right)+4
sqrt2 log left(cos left(fracpi 8right)right)right)-4$$
$$f(3)=frac2 left(pi -sqrt3 log left(sqrt3-1right)+sqrt3 log
left(1+sqrt3right)right)left(sqrt3-1right)
left(1+sqrt3right)-6$$
$$f(4)=frac14 left(pi tan left(fracpi 16right)+pi cot left(fracpi
16right)-8 sin left(fracpi 8right) log left(sin left(frac3
pi 16right)right)+8 cos left(fracpi 8right) log left(cos
left(fracpi 16right)right)-8 cos left(fracpi 8right) log
left(sin left(fracpi 16right)right)+8 sin left(fracpi
8right) log left(cos left(frac3 pi 16right)right)right)-8$$
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
$endgroup$
– user150203
Dec 27 '18 at 8:49
$begingroup$
@DavidG. Your post is very interesting. I need to spend more time with it !
$endgroup$
– Claude Leibovici
Dec 27 '18 at 8:54
$begingroup$
Please do! I look forward to any comments you may have. Thanks again for your post.
$endgroup$
– user150203
Dec 27 '18 at 9:48
$begingroup$
Mathematica calculates the same result
$endgroup$
– stocha
Mar 22 at 19:11
add a comment |
$begingroup$
This does not answer the question as asked in the post.
Consider
$$J_m=int log(1+x^2m),dx$$ One integration by parts gives
$$J_m=x log left(1+x^2 mright)-2mint frac x^2 m+1-1x^2 m+1,dx=x log left(1+x^2 mright)-2mx+2mint fracdxx^2 m+1$$ and
$$int fracdxx^2 m+1=x , _2F_1left(1,frac12 m;1+frac12 m;-x^2 mright)$$ where appears the Gaussian or ordinary hypergeometric function.
So
$$K_m=int_0^a log(1+x^2m),dx=a log left(1+a^2 mright)-2ma+2ma , _2F_1left(1,frac12 m;1+frac12 m;-a^2 mright)$$ and, if $a=1$,
$$I_m=int_0^1 log(1+x^2m),dx= log left(2right)-2m+2m , _2F_1left(1,frac12 m;1+frac12 m;-1right)$$ which can write
$$I_m=log (2)-Phi left(-1,1,1+frac12 mright)$$
where appears the Lerch transcendent function.
Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac12 mright)$ before any simplification
$$f(1)=fracpi 2-2$$
$$f(2)=frac14 left(pi tan left(fracpi 8right)+pi cot left(fracpi
8right)-4 sqrt2 log left(sin left(fracpi 8right)right)+4
sqrt2 log left(cos left(fracpi 8right)right)right)-4$$
$$f(3)=frac2 left(pi -sqrt3 log left(sqrt3-1right)+sqrt3 log
left(1+sqrt3right)right)left(sqrt3-1right)
left(1+sqrt3right)-6$$
$$f(4)=frac14 left(pi tan left(fracpi 16right)+pi cot left(fracpi
16right)-8 sin left(fracpi 8right) log left(sin left(frac3
pi 16right)right)+8 cos left(fracpi 8right) log left(cos
left(fracpi 16right)right)-8 cos left(fracpi 8right) log
left(sin left(fracpi 16right)right)+8 sin left(fracpi
8right) log left(cos left(frac3 pi 16right)right)right)-8$$
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
This does not answer the question as asked in the post.
Consider
$$J_m=int log(1+x^2m),dx$$ One integration by parts gives
$$J_m=x log left(1+x^2 mright)-2mint frac x^2 m+1-1x^2 m+1,dx=x log left(1+x^2 mright)-2mx+2mint fracdxx^2 m+1$$ and
$$int fracdxx^2 m+1=x , _2F_1left(1,frac12 m;1+frac12 m;-x^2 mright)$$ where appears the Gaussian or ordinary hypergeometric function.
So
$$K_m=int_0^a log(1+x^2m),dx=a log left(1+a^2 mright)-2ma+2ma , _2F_1left(1,frac12 m;1+frac12 m;-a^2 mright)$$ and, if $a=1$,
$$I_m=int_0^1 log(1+x^2m),dx= log left(2right)-2m+2m , _2F_1left(1,frac12 m;1+frac12 m;-1right)$$ which can write
$$I_m=log (2)-Phi left(-1,1,1+frac12 mright)$$
where appears the Lerch transcendent function.
Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac12 mright)$ before any simplification
$$f(1)=fracpi 2-2$$
$$f(2)=frac14 left(pi tan left(fracpi 8right)+pi cot left(fracpi
8right)-4 sqrt2 log left(sin left(fracpi 8right)right)+4
sqrt2 log left(cos left(fracpi 8right)right)right)-4$$
$$f(3)=frac2 left(pi -sqrt3 log left(sqrt3-1right)+sqrt3 log
left(1+sqrt3right)right)left(sqrt3-1right)
left(1+sqrt3right)-6$$
$$f(4)=frac14 left(pi tan left(fracpi 16right)+pi cot left(fracpi
16right)-8 sin left(fracpi 8right) log left(sin left(frac3
pi 16right)right)+8 cos left(fracpi 8right) log left(cos
left(fracpi 16right)right)-8 cos left(fracpi 8right) log
left(sin left(fracpi 16right)right)+8 sin left(fracpi
8right) log left(cos left(frac3 pi 16right)right)right)-8$$
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 27 '18 at 8:47
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
$endgroup$
– user150203
Dec 27 '18 at 8:49
$begingroup$
@DavidG. Your post is very interesting. I need to spend more time with it !
$endgroup$
– Claude Leibovici
Dec 27 '18 at 8:54
$begingroup$
Please do! I look forward to any comments you may have. Thanks again for your post.
$endgroup$
– user150203
Dec 27 '18 at 9:48
$begingroup$
Mathematica calculates the same result
$endgroup$
– stocha
Mar 22 at 19:11
add a comment |
$begingroup$
Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
$endgroup$
– user150203
Dec 27 '18 at 8:49
$begingroup$
@DavidG. Your post is very interesting. I need to spend more time with it !
$endgroup$
– Claude Leibovici
Dec 27 '18 at 8:54
$begingroup$
Please do! I look forward to any comments you may have. Thanks again for your post.
$endgroup$
– user150203
Dec 27 '18 at 9:48
$begingroup$
Mathematica calculates the same result
$endgroup$
– stocha
Mar 22 at 19:11
$begingroup$
Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
$endgroup$
– user150203
Dec 27 '18 at 8:49
$begingroup$
Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
$endgroup$
– user150203
Dec 27 '18 at 8:49
$begingroup$
@DavidG. Your post is very interesting. I need to spend more time with it !
$endgroup$
– Claude Leibovici
Dec 27 '18 at 8:54
$begingroup$
@DavidG. Your post is very interesting. I need to spend more time with it !
$endgroup$
– Claude Leibovici
Dec 27 '18 at 8:54
$begingroup$
Please do! I look forward to any comments you may have. Thanks again for your post.
$endgroup$
– user150203
Dec 27 '18 at 9:48
$begingroup$
Please do! I look forward to any comments you may have. Thanks again for your post.
$endgroup$
– user150203
Dec 27 '18 at 9:48
$begingroup$
Mathematica calculates the same result
$endgroup$
– stocha
Mar 22 at 19:11
$begingroup$
Mathematica calculates the same result
$endgroup$
– stocha
Mar 22 at 19:11
add a comment |
$begingroup$
$colorbrowntextbfPreliminary notes.$
Calculation of the required sum looks more complex task than of the
issue integral.
Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.
Denote
$$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
Easy to see that the issue integral is $I_2m.$
$colorbrowntextbfClosed form of the integrals.$
The first step is the integration by parts:
$$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfracx^n-11+x^nmathrm dx,$$
$$I_n = log2 - n + nintlimits_0^1dfracmathrm dx1+x^n.tag2$$
(see also Claude Leibovici).
Then, the substitution
$$x=e^-ttag3$$
gives
$$J_n=intlimits_0^1dfracmathrm dx1+x^n=intlimits_0^inftydfrace^-t,mathrm dt1+e^-nt = sumlimits_k=0^infty(-1)^kintlimits_0^infty e^-(kn+1)t,mathrm dt=sumlimits_k=0^inftydfrac(-1)^kkn+1 ,$$
$$J_n = intlimits_0^1dfracmathrm dx1+x^n= dfrac12nleft(psi_0left(dfracn+12nright) - psi_0left(dfrac12nright)right),tag4$$
where $psi_0(x)$ is the polygamma function.
Using $(2),(4),$ easy to get the expression for the OP integral in the form of
$$boxedI_2m = log2 - 2m +dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right).tag5$$
$colorbrowntextbfTesting of the solution.$
Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.
Immediate calculations allow to check obtained expressions in the simple cases
beginalign
&J_1 = intlimits_0^1dfracmathrm dx1+x = log 2,\[4pt]
&J_2 = intlimits_0^1dfracmathrm dx1+x^2 = arctan 1 = dfracpi4,\[4pt]
&J_3 = intlimits_0^1dfracmathrm dx1+x^3
= dfrac16intlimits_0^1left(dfrac21+x - dfrac2x-11-x+x^2 + dfrac31-x+x^2right),mathrm dx\
&= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1sqrt3arctandfrac2x-1sqrt3right)bigg|_0^1
= dfracpi3sqrt3 + frac13 log2.\[4pt]
endalign
More hard cases are
beginalign
&J_4 = intlimits_0^1dfracmathrm dx1+x^4
= dfracsqrt28left(logdfracx^2 + sqrt2 x + 1x^2 - sqrt2 x + 1
- 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
&= dfracsqrt28left(logdfrac(x^2 + sqrt2 x + 1)^21+x^4 + 2arctandfracxsqrt21-x^2right)bigg|_0^1
= dfracsqrt28(pi+log(3+2sqrt2)),\[4pt]
&J_4=dfrac18Bigg(dfrac12sqrtdfrac2-sqrt22+sqrt2pi+dfrac12sqrtdfrac2+sqrt22-sqrt2pi
-2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
&+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
&=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
+sqrt2logdfrac2+sqrt22-sqrt2Bigg)
= dfracsqrt28(pi+log(3+2sqrt2)),
endalign
and $n>4.$
Numeric calculations of the issue integral $I_2m$ and its closed form $(5)$ also confirm the correctness of the closed form.
$colorbrowntextbfThe alternative approach.$
The alternative approach is considered in OP. Let us repeat it with some differences.
Polynomial factorization can be presented in the form of
beginalign
&1+x^2m = prodlimits_j=0^2m-1large left(x-e^frac2j+12mpi iright)
= prodlimits_j=0^m-1left(x^2-2xcosfrac2j+12mpi+1right),
endalign
so
beginalign
&I_2m = intlimits_0^1 ln(1+x^2m),mathrm dx
= sumlimits_j=0^m-1 T_j,quadtextwhere\[4pt]
&T_j = intlimits_0^1 lnleft(x^2-2xcosfrac2j+12mpi+1right),mathrm dx,\[4pt]
&T_j = xlnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1
- 2intlimits_0^1 dfracxleft(x-cosfrac2j+12mpiright)x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright)
- 2-2intlimits_0^1 dfracxcosfrac2j+12mpi-1x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- 2cosfrac2j+12mpi intlimits_0^1 dfracx-cosfrac2j+12mpix^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& + 2intlimits_0^1 dfracsin^2frac2j+12mpileft(x-cosfrac2j+12mpiright)^2+sin^2frac2j+12mpi,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1\[4pt]
& + 2sinfrac2j+12mpi arctandfracx-cosfrac2j+12mpisinfrac2j+12mpibigg|_0^1\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
left(arctandfrac1-cosfrac2j+12mpisinfrac2j+12mpi
+ arctandfraccosfrac2j+12mpisinfrac2j+12mpiright)\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
arctandfracsinfrac2j+12mpi
sin^2frac2j+12mpi - left(1 - cosfrac2j+12mpiright)cosfrac2j+12mpi\[4pt]
& = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + 2sinfrac2j+12mpiarctancotfrac2j+14mpi,\[4pt]
&T_j = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + dfrac2(m-j)-12mpisinfrac2j+12mpi.
endalign
Taking in account that
beginalign
&sumlimits_j=0^m-1cosfrac2j+12mpi
= Re sumlimits_j=0^m-1e^frac2j+12mpi i
= Re large dfrac1-e^pi i1-e^fracpimie^fracpi2mi
=Redfrac isinfracpi2m = 0,\[4pt]
&sumlimits_j=0^m-1sinfrac2j+12mpi
= Im sumlimits_j=0^m-1e^frac2j+12mpi i
=Imdfracisinfracpi2m = dfrac1sinfracpi2m,\[4pt]
&sumlimits_j=0^m-1frac2j+12mpisinfrac2j+12mpi
= dfracpi2sinfracpi2m
endalign
(see also Wolfram Alpha calculations),
one can get
$$boxedI_2m = m(ln2-2) + dfracpi2cscfracpi2m
+ sumlimits_j=0^m-1left(1pmcosfrac2j+12mpiright)
logleft(1pmcosfrac2j+12mpiright).tag6$$
$colorbrowntextbfClosed form for the sum.$
From $(5)-(6)$ should
$$colorgreenboxedsumlimits_j=0^m-1left(1pm c_jright)
logleft(1pm c_jright)
= dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right) - (m-1)ln2 - dfracpi2cscfracpi2m
,tag7$$
where
$$c_j = cosfrac2j+12mpi.$$
Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".
Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
$endgroup$
1
$begingroup$
Very nice presentation, already before the alternate approach (+1).
$endgroup$
– Markus Scheuer
yesterday
$begingroup$
@MarkusScheuer Thanks, you are welcome!
$endgroup$
– Yuri Negometyanov
yesterday
add a comment |
$begingroup$
$colorbrowntextbfPreliminary notes.$
Calculation of the required sum looks more complex task than of the
issue integral.
Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.
Denote
$$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
Easy to see that the issue integral is $I_2m.$
$colorbrowntextbfClosed form of the integrals.$
The first step is the integration by parts:
$$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfracx^n-11+x^nmathrm dx,$$
$$I_n = log2 - n + nintlimits_0^1dfracmathrm dx1+x^n.tag2$$
(see also Claude Leibovici).
Then, the substitution
$$x=e^-ttag3$$
gives
$$J_n=intlimits_0^1dfracmathrm dx1+x^n=intlimits_0^inftydfrace^-t,mathrm dt1+e^-nt = sumlimits_k=0^infty(-1)^kintlimits_0^infty e^-(kn+1)t,mathrm dt=sumlimits_k=0^inftydfrac(-1)^kkn+1 ,$$
$$J_n = intlimits_0^1dfracmathrm dx1+x^n= dfrac12nleft(psi_0left(dfracn+12nright) - psi_0left(dfrac12nright)right),tag4$$
where $psi_0(x)$ is the polygamma function.
Using $(2),(4),$ easy to get the expression for the OP integral in the form of
$$boxedI_2m = log2 - 2m +dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right).tag5$$
$colorbrowntextbfTesting of the solution.$
Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.
Immediate calculations allow to check obtained expressions in the simple cases
beginalign
&J_1 = intlimits_0^1dfracmathrm dx1+x = log 2,\[4pt]
&J_2 = intlimits_0^1dfracmathrm dx1+x^2 = arctan 1 = dfracpi4,\[4pt]
&J_3 = intlimits_0^1dfracmathrm dx1+x^3
= dfrac16intlimits_0^1left(dfrac21+x - dfrac2x-11-x+x^2 + dfrac31-x+x^2right),mathrm dx\
&= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1sqrt3arctandfrac2x-1sqrt3right)bigg|_0^1
= dfracpi3sqrt3 + frac13 log2.\[4pt]
endalign
More hard cases are
beginalign
&J_4 = intlimits_0^1dfracmathrm dx1+x^4
= dfracsqrt28left(logdfracx^2 + sqrt2 x + 1x^2 - sqrt2 x + 1
- 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
&= dfracsqrt28left(logdfrac(x^2 + sqrt2 x + 1)^21+x^4 + 2arctandfracxsqrt21-x^2right)bigg|_0^1
= dfracsqrt28(pi+log(3+2sqrt2)),\[4pt]
&J_4=dfrac18Bigg(dfrac12sqrtdfrac2-sqrt22+sqrt2pi+dfrac12sqrtdfrac2+sqrt22-sqrt2pi
-2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
&+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
&=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
+sqrt2logdfrac2+sqrt22-sqrt2Bigg)
= dfracsqrt28(pi+log(3+2sqrt2)),
endalign
and $n>4.$
Numeric calculations of the issue integral $I_2m$ and its closed form $(5)$ also confirm the correctness of the closed form.
$colorbrowntextbfThe alternative approach.$
The alternative approach is considered in OP. Let us repeat it with some differences.
Polynomial factorization can be presented in the form of
beginalign
&1+x^2m = prodlimits_j=0^2m-1large left(x-e^frac2j+12mpi iright)
= prodlimits_j=0^m-1left(x^2-2xcosfrac2j+12mpi+1right),
endalign
so
beginalign
&I_2m = intlimits_0^1 ln(1+x^2m),mathrm dx
= sumlimits_j=0^m-1 T_j,quadtextwhere\[4pt]
&T_j = intlimits_0^1 lnleft(x^2-2xcosfrac2j+12mpi+1right),mathrm dx,\[4pt]
&T_j = xlnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1
- 2intlimits_0^1 dfracxleft(x-cosfrac2j+12mpiright)x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright)
- 2-2intlimits_0^1 dfracxcosfrac2j+12mpi-1x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- 2cosfrac2j+12mpi intlimits_0^1 dfracx-cosfrac2j+12mpix^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& + 2intlimits_0^1 dfracsin^2frac2j+12mpileft(x-cosfrac2j+12mpiright)^2+sin^2frac2j+12mpi,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1\[4pt]
& + 2sinfrac2j+12mpi arctandfracx-cosfrac2j+12mpisinfrac2j+12mpibigg|_0^1\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
left(arctandfrac1-cosfrac2j+12mpisinfrac2j+12mpi
+ arctandfraccosfrac2j+12mpisinfrac2j+12mpiright)\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
arctandfracsinfrac2j+12mpi
sin^2frac2j+12mpi - left(1 - cosfrac2j+12mpiright)cosfrac2j+12mpi\[4pt]
& = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + 2sinfrac2j+12mpiarctancotfrac2j+14mpi,\[4pt]
&T_j = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + dfrac2(m-j)-12mpisinfrac2j+12mpi.
endalign
Taking in account that
beginalign
&sumlimits_j=0^m-1cosfrac2j+12mpi
= Re sumlimits_j=0^m-1e^frac2j+12mpi i
= Re large dfrac1-e^pi i1-e^fracpimie^fracpi2mi
=Redfrac isinfracpi2m = 0,\[4pt]
&sumlimits_j=0^m-1sinfrac2j+12mpi
= Im sumlimits_j=0^m-1e^frac2j+12mpi i
=Imdfracisinfracpi2m = dfrac1sinfracpi2m,\[4pt]
&sumlimits_j=0^m-1frac2j+12mpisinfrac2j+12mpi
= dfracpi2sinfracpi2m
endalign
(see also Wolfram Alpha calculations),
one can get
$$boxedI_2m = m(ln2-2) + dfracpi2cscfracpi2m
+ sumlimits_j=0^m-1left(1pmcosfrac2j+12mpiright)
logleft(1pmcosfrac2j+12mpiright).tag6$$
$colorbrowntextbfClosed form for the sum.$
From $(5)-(6)$ should
$$colorgreenboxedsumlimits_j=0^m-1left(1pm c_jright)
logleft(1pm c_jright)
= dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right) - (m-1)ln2 - dfracpi2cscfracpi2m
,tag7$$
where
$$c_j = cosfrac2j+12mpi.$$
Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".
Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
$endgroup$
1
$begingroup$
Very nice presentation, already before the alternate approach (+1).
$endgroup$
– Markus Scheuer
yesterday
$begingroup$
@MarkusScheuer Thanks, you are welcome!
$endgroup$
– Yuri Negometyanov
yesterday
add a comment |
$begingroup$
$colorbrowntextbfPreliminary notes.$
Calculation of the required sum looks more complex task than of the
issue integral.
Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.
Denote
$$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
Easy to see that the issue integral is $I_2m.$
$colorbrowntextbfClosed form of the integrals.$
The first step is the integration by parts:
$$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfracx^n-11+x^nmathrm dx,$$
$$I_n = log2 - n + nintlimits_0^1dfracmathrm dx1+x^n.tag2$$
(see also Claude Leibovici).
Then, the substitution
$$x=e^-ttag3$$
gives
$$J_n=intlimits_0^1dfracmathrm dx1+x^n=intlimits_0^inftydfrace^-t,mathrm dt1+e^-nt = sumlimits_k=0^infty(-1)^kintlimits_0^infty e^-(kn+1)t,mathrm dt=sumlimits_k=0^inftydfrac(-1)^kkn+1 ,$$
$$J_n = intlimits_0^1dfracmathrm dx1+x^n= dfrac12nleft(psi_0left(dfracn+12nright) - psi_0left(dfrac12nright)right),tag4$$
where $psi_0(x)$ is the polygamma function.
Using $(2),(4),$ easy to get the expression for the OP integral in the form of
$$boxedI_2m = log2 - 2m +dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right).tag5$$
$colorbrowntextbfTesting of the solution.$
Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.
Immediate calculations allow to check obtained expressions in the simple cases
beginalign
&J_1 = intlimits_0^1dfracmathrm dx1+x = log 2,\[4pt]
&J_2 = intlimits_0^1dfracmathrm dx1+x^2 = arctan 1 = dfracpi4,\[4pt]
&J_3 = intlimits_0^1dfracmathrm dx1+x^3
= dfrac16intlimits_0^1left(dfrac21+x - dfrac2x-11-x+x^2 + dfrac31-x+x^2right),mathrm dx\
&= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1sqrt3arctandfrac2x-1sqrt3right)bigg|_0^1
= dfracpi3sqrt3 + frac13 log2.\[4pt]
endalign
More hard cases are
beginalign
&J_4 = intlimits_0^1dfracmathrm dx1+x^4
= dfracsqrt28left(logdfracx^2 + sqrt2 x + 1x^2 - sqrt2 x + 1
- 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
&= dfracsqrt28left(logdfrac(x^2 + sqrt2 x + 1)^21+x^4 + 2arctandfracxsqrt21-x^2right)bigg|_0^1
= dfracsqrt28(pi+log(3+2sqrt2)),\[4pt]
&J_4=dfrac18Bigg(dfrac12sqrtdfrac2-sqrt22+sqrt2pi+dfrac12sqrtdfrac2+sqrt22-sqrt2pi
-2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
&+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
&=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
+sqrt2logdfrac2+sqrt22-sqrt2Bigg)
= dfracsqrt28(pi+log(3+2sqrt2)),
endalign
and $n>4.$
Numeric calculations of the issue integral $I_2m$ and its closed form $(5)$ also confirm the correctness of the closed form.
$colorbrowntextbfThe alternative approach.$
The alternative approach is considered in OP. Let us repeat it with some differences.
Polynomial factorization can be presented in the form of
beginalign
&1+x^2m = prodlimits_j=0^2m-1large left(x-e^frac2j+12mpi iright)
= prodlimits_j=0^m-1left(x^2-2xcosfrac2j+12mpi+1right),
endalign
so
beginalign
&I_2m = intlimits_0^1 ln(1+x^2m),mathrm dx
= sumlimits_j=0^m-1 T_j,quadtextwhere\[4pt]
&T_j = intlimits_0^1 lnleft(x^2-2xcosfrac2j+12mpi+1right),mathrm dx,\[4pt]
&T_j = xlnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1
- 2intlimits_0^1 dfracxleft(x-cosfrac2j+12mpiright)x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright)
- 2-2intlimits_0^1 dfracxcosfrac2j+12mpi-1x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- 2cosfrac2j+12mpi intlimits_0^1 dfracx-cosfrac2j+12mpix^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& + 2intlimits_0^1 dfracsin^2frac2j+12mpileft(x-cosfrac2j+12mpiright)^2+sin^2frac2j+12mpi,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1\[4pt]
& + 2sinfrac2j+12mpi arctandfracx-cosfrac2j+12mpisinfrac2j+12mpibigg|_0^1\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
left(arctandfrac1-cosfrac2j+12mpisinfrac2j+12mpi
+ arctandfraccosfrac2j+12mpisinfrac2j+12mpiright)\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
arctandfracsinfrac2j+12mpi
sin^2frac2j+12mpi - left(1 - cosfrac2j+12mpiright)cosfrac2j+12mpi\[4pt]
& = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + 2sinfrac2j+12mpiarctancotfrac2j+14mpi,\[4pt]
&T_j = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + dfrac2(m-j)-12mpisinfrac2j+12mpi.
endalign
Taking in account that
beginalign
&sumlimits_j=0^m-1cosfrac2j+12mpi
= Re sumlimits_j=0^m-1e^frac2j+12mpi i
= Re large dfrac1-e^pi i1-e^fracpimie^fracpi2mi
=Redfrac isinfracpi2m = 0,\[4pt]
&sumlimits_j=0^m-1sinfrac2j+12mpi
= Im sumlimits_j=0^m-1e^frac2j+12mpi i
=Imdfracisinfracpi2m = dfrac1sinfracpi2m,\[4pt]
&sumlimits_j=0^m-1frac2j+12mpisinfrac2j+12mpi
= dfracpi2sinfracpi2m
endalign
(see also Wolfram Alpha calculations),
one can get
$$boxedI_2m = m(ln2-2) + dfracpi2cscfracpi2m
+ sumlimits_j=0^m-1left(1pmcosfrac2j+12mpiright)
logleft(1pmcosfrac2j+12mpiright).tag6$$
$colorbrowntextbfClosed form for the sum.$
From $(5)-(6)$ should
$$colorgreenboxedsumlimits_j=0^m-1left(1pm c_jright)
logleft(1pm c_jright)
= dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right) - (m-1)ln2 - dfracpi2cscfracpi2m
,tag7$$
where
$$c_j = cosfrac2j+12mpi.$$
Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".
Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
$endgroup$
$colorbrowntextbfPreliminary notes.$
Calculation of the required sum looks more complex task than of the
issue integral.
Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.
Denote
$$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
Easy to see that the issue integral is $I_2m.$
$colorbrowntextbfClosed form of the integrals.$
The first step is the integration by parts:
$$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfracx^n-11+x^nmathrm dx,$$
$$I_n = log2 - n + nintlimits_0^1dfracmathrm dx1+x^n.tag2$$
(see also Claude Leibovici).
Then, the substitution
$$x=e^-ttag3$$
gives
$$J_n=intlimits_0^1dfracmathrm dx1+x^n=intlimits_0^inftydfrace^-t,mathrm dt1+e^-nt = sumlimits_k=0^infty(-1)^kintlimits_0^infty e^-(kn+1)t,mathrm dt=sumlimits_k=0^inftydfrac(-1)^kkn+1 ,$$
$$J_n = intlimits_0^1dfracmathrm dx1+x^n= dfrac12nleft(psi_0left(dfracn+12nright) - psi_0left(dfrac12nright)right),tag4$$
where $psi_0(x)$ is the polygamma function.
Using $(2),(4),$ easy to get the expression for the OP integral in the form of
$$boxedI_2m = log2 - 2m +dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right).tag5$$
$colorbrowntextbfTesting of the solution.$
Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.
Immediate calculations allow to check obtained expressions in the simple cases
beginalign
&J_1 = intlimits_0^1dfracmathrm dx1+x = log 2,\[4pt]
&J_2 = intlimits_0^1dfracmathrm dx1+x^2 = arctan 1 = dfracpi4,\[4pt]
&J_3 = intlimits_0^1dfracmathrm dx1+x^3
= dfrac16intlimits_0^1left(dfrac21+x - dfrac2x-11-x+x^2 + dfrac31-x+x^2right),mathrm dx\
&= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1sqrt3arctandfrac2x-1sqrt3right)bigg|_0^1
= dfracpi3sqrt3 + frac13 log2.\[4pt]
endalign
More hard cases are
beginalign
&J_4 = intlimits_0^1dfracmathrm dx1+x^4
= dfracsqrt28left(logdfracx^2 + sqrt2 x + 1x^2 - sqrt2 x + 1
- 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
&= dfracsqrt28left(logdfrac(x^2 + sqrt2 x + 1)^21+x^4 + 2arctandfracxsqrt21-x^2right)bigg|_0^1
= dfracsqrt28(pi+log(3+2sqrt2)),\[4pt]
&J_4=dfrac18Bigg(dfrac12sqrtdfrac2-sqrt22+sqrt2pi+dfrac12sqrtdfrac2+sqrt22-sqrt2pi
-2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
&+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
&=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
+sqrt2logdfrac2+sqrt22-sqrt2Bigg)
= dfracsqrt28(pi+log(3+2sqrt2)),
endalign
and $n>4.$
Numeric calculations of the issue integral $I_2m$ and its closed form $(5)$ also confirm the correctness of the closed form.
$colorbrowntextbfThe alternative approach.$
The alternative approach is considered in OP. Let us repeat it with some differences.
Polynomial factorization can be presented in the form of
beginalign
&1+x^2m = prodlimits_j=0^2m-1large left(x-e^frac2j+12mpi iright)
= prodlimits_j=0^m-1left(x^2-2xcosfrac2j+12mpi+1right),
endalign
so
beginalign
&I_2m = intlimits_0^1 ln(1+x^2m),mathrm dx
= sumlimits_j=0^m-1 T_j,quadtextwhere\[4pt]
&T_j = intlimits_0^1 lnleft(x^2-2xcosfrac2j+12mpi+1right),mathrm dx,\[4pt]
&T_j = xlnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1
- 2intlimits_0^1 dfracxleft(x-cosfrac2j+12mpiright)x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright)
- 2-2intlimits_0^1 dfracxcosfrac2j+12mpi-1x^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- 2cosfrac2j+12mpi intlimits_0^1 dfracx-cosfrac2j+12mpix^2-2xcosfrac2j+12mpi+1,mathrm dx\[4pt]
& + 2intlimits_0^1 dfracsin^2frac2j+12mpileft(x-cosfrac2j+12mpiright)^2+sin^2frac2j+12mpi,mathrm dx\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(x^2-2xcosfrac2j+12mpi+1right)bigg|_0^1\[4pt]
& + 2sinfrac2j+12mpi arctandfracx-cosfrac2j+12mpisinfrac2j+12mpibigg|_0^1\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
left(arctandfrac1-cosfrac2j+12mpisinfrac2j+12mpi
+ arctandfraccosfrac2j+12mpisinfrac2j+12mpiright)\[4pt]
& = lnleft(2-2cosfrac2j+12mpiright) - 2
- cosfrac2j+12mpi lnleft(2-2cosfrac2j+12mpiright)\[4pt]
& + 2sinfrac2j+12mpi
arctandfracsinfrac2j+12mpi
sin^2frac2j+12mpi - left(1 - cosfrac2j+12mpiright)cosfrac2j+12mpi\[4pt]
& = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + 2sinfrac2j+12mpiarctancotfrac2j+14mpi,\[4pt]
&T_j = left(1-cosfrac2j+12mpiright)lnleft(2-2cosfrac2j+12mpiright) - 2 + dfrac2(m-j)-12mpisinfrac2j+12mpi.
endalign
Taking in account that
beginalign
&sumlimits_j=0^m-1cosfrac2j+12mpi
= Re sumlimits_j=0^m-1e^frac2j+12mpi i
= Re large dfrac1-e^pi i1-e^fracpimie^fracpi2mi
=Redfrac isinfracpi2m = 0,\[4pt]
&sumlimits_j=0^m-1sinfrac2j+12mpi
= Im sumlimits_j=0^m-1e^frac2j+12mpi i
=Imdfracisinfracpi2m = dfrac1sinfracpi2m,\[4pt]
&sumlimits_j=0^m-1frac2j+12mpisinfrac2j+12mpi
= dfracpi2sinfracpi2m
endalign
(see also Wolfram Alpha calculations),
one can get
$$boxedI_2m = m(ln2-2) + dfracpi2cscfracpi2m
+ sumlimits_j=0^m-1left(1pmcosfrac2j+12mpiright)
logleft(1pmcosfrac2j+12mpiright).tag6$$
$colorbrowntextbfClosed form for the sum.$
From $(5)-(6)$ should
$$colorgreenboxedsumlimits_j=0^m-1left(1pm c_jright)
logleft(1pm c_jright)
= dfrac12left(psi_0left(dfrac2m+14mright) - psi_0left(dfrac14mright)right) - (m-1)ln2 - dfracpi2cscfracpi2m
,tag7$$
where
$$c_j = cosfrac2j+12mpi.$$
Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".
Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
edited yesterday
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
answered Mar 26 at 2:16
Yuri NegometyanovYuri Negometyanov
12.2k1729
12.2k1729
1
$begingroup$
Very nice presentation, already before the alternate approach (+1).
$endgroup$
– Markus Scheuer
yesterday
$begingroup$
@MarkusScheuer Thanks, you are welcome!
$endgroup$
– Yuri Negometyanov
yesterday
add a comment |
1
$begingroup$
Very nice presentation, already before the alternate approach (+1).
$endgroup$
– Markus Scheuer
yesterday
$begingroup$
@MarkusScheuer Thanks, you are welcome!
$endgroup$
– Yuri Negometyanov
yesterday
1
1
$begingroup$
Very nice presentation, already before the alternate approach (+1).
$endgroup$
– Markus Scheuer
yesterday
$begingroup$
Very nice presentation, already before the alternate approach (+1).
$endgroup$
– Markus Scheuer
yesterday
$begingroup$
@MarkusScheuer Thanks, you are welcome!
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@MarkusScheuer Thanks, you are welcome!
$endgroup$
– Yuri Negometyanov
yesterday
add a comment |
$begingroup$
I did it!
I actually have no idea whether or not this works, but this is how I did it.
$ninBbb N$
Define the sequence $r_k^(n)_k=1^k=n$ such that
$$x^n+1=prod_k=1^nbig(x-r^(n)_kbig)$$
We then know that $$r_k^(n)=expbigg[fracipin(2k-1)bigg]$$
Then we define
$$S_n=r_k^(n):kin[1,n]capBbb N$$
So we have that
$$frac1x^n+1=prod_rin S_nfrac1x-r=prod_k=1^nfrac1x-r_k^(n)$$
Then we assume that we can write
$$prod_rin S_nfrac1x-r=sum_rin S_nfracb(r)x-r$$
Multiplying both sides by $prod_ain S_n(x-a)$,
$$1=sum_rin S_nb(r)prod_ain S_n\ aneq r(x-a)$$
So for any $omegain S_n$,
$$1=b(omega)prod_ain S_n\ aneq omega(omega-a)$$
$$b(omega)=prod_ain S_n\ aneq omegafrac1omega-a$$
$$b(r_k^(n))=prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we know that
$$I_n=int_0^1fracmathrmdx1+x^n=sum_k=1^nb(r_k^(n))int_0^1fracmathrmdxx-r_k^(n)$$
$$I_n=sum_k=1^nb(r_k^(n))logbigg|fracr_k^(n)-1r_k^(n)bigg|$$
$$I_n=sum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we have
$$int_0^1log(1+x^n)mathrmdx=log2-n+nsum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
along with a plethora of other identities...
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
$endgroup$
1
$begingroup$
Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
$endgroup$
– user150203
Dec 31 '18 at 6:16
$begingroup$
I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
$endgroup$
– user150203
Dec 31 '18 at 9:47
$begingroup$
Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
$endgroup$
– clathratus
Dec 31 '18 at 20:28
add a comment |
$begingroup$
I did it!
I actually have no idea whether or not this works, but this is how I did it.
$ninBbb N$
Define the sequence $r_k^(n)_k=1^k=n$ such that
$$x^n+1=prod_k=1^nbig(x-r^(n)_kbig)$$
We then know that $$r_k^(n)=expbigg[fracipin(2k-1)bigg]$$
Then we define
$$S_n=r_k^(n):kin[1,n]capBbb N$$
So we have that
$$frac1x^n+1=prod_rin S_nfrac1x-r=prod_k=1^nfrac1x-r_k^(n)$$
Then we assume that we can write
$$prod_rin S_nfrac1x-r=sum_rin S_nfracb(r)x-r$$
Multiplying both sides by $prod_ain S_n(x-a)$,
$$1=sum_rin S_nb(r)prod_ain S_n\ aneq r(x-a)$$
So for any $omegain S_n$,
$$1=b(omega)prod_ain S_n\ aneq omega(omega-a)$$
$$b(omega)=prod_ain S_n\ aneq omegafrac1omega-a$$
$$b(r_k^(n))=prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we know that
$$I_n=int_0^1fracmathrmdx1+x^n=sum_k=1^nb(r_k^(n))int_0^1fracmathrmdxx-r_k^(n)$$
$$I_n=sum_k=1^nb(r_k^(n))logbigg|fracr_k^(n)-1r_k^(n)bigg|$$
$$I_n=sum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we have
$$int_0^1log(1+x^n)mathrmdx=log2-n+nsum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
along with a plethora of other identities...
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
$endgroup$
1
$begingroup$
Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
$endgroup$
– user150203
Dec 31 '18 at 6:16
$begingroup$
I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
$endgroup$
– user150203
Dec 31 '18 at 9:47
$begingroup$
Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
$endgroup$
– clathratus
Dec 31 '18 at 20:28
add a comment |
$begingroup$
I did it!
I actually have no idea whether or not this works, but this is how I did it.
$ninBbb N$
Define the sequence $r_k^(n)_k=1^k=n$ such that
$$x^n+1=prod_k=1^nbig(x-r^(n)_kbig)$$
We then know that $$r_k^(n)=expbigg[fracipin(2k-1)bigg]$$
Then we define
$$S_n=r_k^(n):kin[1,n]capBbb N$$
So we have that
$$frac1x^n+1=prod_rin S_nfrac1x-r=prod_k=1^nfrac1x-r_k^(n)$$
Then we assume that we can write
$$prod_rin S_nfrac1x-r=sum_rin S_nfracb(r)x-r$$
Multiplying both sides by $prod_ain S_n(x-a)$,
$$1=sum_rin S_nb(r)prod_ain S_n\ aneq r(x-a)$$
So for any $omegain S_n$,
$$1=b(omega)prod_ain S_n\ aneq omega(omega-a)$$
$$b(omega)=prod_ain S_n\ aneq omegafrac1omega-a$$
$$b(r_k^(n))=prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we know that
$$I_n=int_0^1fracmathrmdx1+x^n=sum_k=1^nb(r_k^(n))int_0^1fracmathrmdxx-r_k^(n)$$
$$I_n=sum_k=1^nb(r_k^(n))logbigg|fracr_k^(n)-1r_k^(n)bigg|$$
$$I_n=sum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we have
$$int_0^1log(1+x^n)mathrmdx=log2-n+nsum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
along with a plethora of other identities...
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
$endgroup$
I did it!
I actually have no idea whether or not this works, but this is how I did it.
$ninBbb N$
Define the sequence $r_k^(n)_k=1^k=n$ such that
$$x^n+1=prod_k=1^nbig(x-r^(n)_kbig)$$
We then know that $$r_k^(n)=expbigg[fracipin(2k-1)bigg]$$
Then we define
$$S_n=r_k^(n):kin[1,n]capBbb N$$
So we have that
$$frac1x^n+1=prod_rin S_nfrac1x-r=prod_k=1^nfrac1x-r_k^(n)$$
Then we assume that we can write
$$prod_rin S_nfrac1x-r=sum_rin S_nfracb(r)x-r$$
Multiplying both sides by $prod_ain S_n(x-a)$,
$$1=sum_rin S_nb(r)prod_ain S_n\ aneq r(x-a)$$
So for any $omegain S_n$,
$$1=b(omega)prod_ain S_n\ aneq omega(omega-a)$$
$$b(omega)=prod_ain S_n\ aneq omegafrac1omega-a$$
$$b(r_k^(n))=prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we know that
$$I_n=int_0^1fracmathrmdx1+x^n=sum_k=1^nb(r_k^(n))int_0^1fracmathrmdxx-r_k^(n)$$
$$I_n=sum_k=1^nb(r_k^(n))logbigg|fracr_k^(n)-1r_k^(n)bigg|$$
$$I_n=sum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
So we have
$$int_0^1log(1+x^n)mathrmdx=log2-n+nsum_k=1^nlogbigg|fracr_k^(n)-1r_k^(n)bigg|prod_p=1\ pneq k^nfrac1r_k^(n)-r_p^(n)$$
along with a plethora of other identities...
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
edited Jan 11 at 21:56
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
answered Dec 31 '18 at 5:52
clathratusclathratus
5,0011438
5,0011438
1
$begingroup$
Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
$endgroup$
– user150203
Dec 31 '18 at 6:16
$begingroup$
I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
$endgroup$
– user150203
Dec 31 '18 at 9:47
$begingroup$
Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
$endgroup$
– clathratus
Dec 31 '18 at 20:28
add a comment |
1
$begingroup$
Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
$endgroup$
– user150203
Dec 31 '18 at 6:16
$begingroup$
I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
$endgroup$
– user150203
Dec 31 '18 at 9:47
$begingroup$
Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
$endgroup$
– clathratus
Dec 31 '18 at 20:28
1
1
$begingroup$
Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
$endgroup$
– user150203
Dec 31 '18 at 6:16
$begingroup$
Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
$endgroup$
– user150203
Dec 31 '18 at 6:16
$begingroup$
I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
$endgroup$
– user150203
Dec 31 '18 at 9:47
$begingroup$
I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
$endgroup$
– user150203
Dec 31 '18 at 9:47
$begingroup$
Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
$endgroup$
– clathratus
Dec 31 '18 at 20:28
$begingroup$
Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
$endgroup$
– clathratus
Dec 31 '18 at 20:28
add a comment |
$begingroup$
Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^(n)$ as last time, we apply the $logprod_ia_i=sum_ilog a_i$ property to see that
$$log(1+x^n)=logprod_k=1^n(x-r_k^(n))=sum_k=1^nlog(x-r_k^(n))$$
So
$$I_n=int_0^1log(1+x^n)mathrm dx=sum_k=1^nint_0^1log(x-r_k^(n))mathrm dx$$
This last integral boils down to
$$beginalign
int_0^1log(x-a)mathrm dx=&alogfraca1+a+log(1-a)-1\
=&logfraca^a(1-a)e(1+a)^a
endalign$$
So
$$I_n=sum_rin S_nlogfracr^r(1-r)e(1+r)^r$$
And you know how I love product representations, so we again use $logprod_ia_i=sum_ilog a_i$ to see that
$$
I_n=logprod_rin S_nfracr^r(1-r)e(1+r)^r\
prod_rin S_nfracr^r(1-r)(1+r)^r=exp(n+I_n)
$$
Which I just think is really neat.
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
$endgroup$
add a comment |
$begingroup$
Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^(n)$ as last time, we apply the $logprod_ia_i=sum_ilog a_i$ property to see that
$$log(1+x^n)=logprod_k=1^n(x-r_k^(n))=sum_k=1^nlog(x-r_k^(n))$$
So
$$I_n=int_0^1log(1+x^n)mathrm dx=sum_k=1^nint_0^1log(x-r_k^(n))mathrm dx$$
This last integral boils down to
$$beginalign
int_0^1log(x-a)mathrm dx=&alogfraca1+a+log(1-a)-1\
=&logfraca^a(1-a)e(1+a)^a
endalign$$
So
$$I_n=sum_rin S_nlogfracr^r(1-r)e(1+r)^r$$
And you know how I love product representations, so we again use $logprod_ia_i=sum_ilog a_i$ to see that
$$
I_n=logprod_rin S_nfracr^r(1-r)e(1+r)^r\
prod_rin S_nfracr^r(1-r)(1+r)^r=exp(n+I_n)
$$
Which I just think is really neat.
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
$endgroup$
add a comment |
$begingroup$
Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^(n)$ as last time, we apply the $logprod_ia_i=sum_ilog a_i$ property to see that
$$log(1+x^n)=logprod_k=1^n(x-r_k^(n))=sum_k=1^nlog(x-r_k^(n))$$
So
$$I_n=int_0^1log(1+x^n)mathrm dx=sum_k=1^nint_0^1log(x-r_k^(n))mathrm dx$$
This last integral boils down to
$$beginalign
int_0^1log(x-a)mathrm dx=&alogfraca1+a+log(1-a)-1\
=&logfraca^a(1-a)e(1+a)^a
endalign$$
So
$$I_n=sum_rin S_nlogfracr^r(1-r)e(1+r)^r$$
And you know how I love product representations, so we again use $logprod_ia_i=sum_ilog a_i$ to see that
$$
I_n=logprod_rin S_nfracr^r(1-r)e(1+r)^r\
prod_rin S_nfracr^r(1-r)(1+r)^r=exp(n+I_n)
$$
Which I just think is really neat.
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
$endgroup$
Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^(n)$ as last time, we apply the $logprod_ia_i=sum_ilog a_i$ property to see that
$$log(1+x^n)=logprod_k=1^n(x-r_k^(n))=sum_k=1^nlog(x-r_k^(n))$$
So
$$I_n=int_0^1log(1+x^n)mathrm dx=sum_k=1^nint_0^1log(x-r_k^(n))mathrm dx$$
This last integral boils down to
$$beginalign
int_0^1log(x-a)mathrm dx=&alogfraca1+a+log(1-a)-1\
=&logfraca^a(1-a)e(1+a)^a
endalign$$
So
$$I_n=sum_rin S_nlogfracr^r(1-r)e(1+r)^r$$
And you know how I love product representations, so we again use $logprod_ia_i=sum_ilog a_i$ to see that
$$
I_n=logprod_rin S_nfracr^r(1-r)e(1+r)^r\
prod_rin S_nfracr^r(1-r)(1+r)^r=exp(n+I_n)
$$
Which I just think is really neat.
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
answered Jan 3 at 23:58
clathratusclathratus
5,0011438
5,0011438
add a comment |
add a comment |
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$begingroup$
@Mason - Yes, thanks for the pickup, I will edit now
$endgroup$
– user150203
Dec 27 '18 at 4:46
$begingroup$
And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
$endgroup$
– Mason
Dec 27 '18 at 4:58
$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
$endgroup$
– user150203
Dec 27 '18 at 5:02
1
$begingroup$
Thanks for the shout-out David ;) I'll try to answer the question
$endgroup$
– clathratus
Dec 27 '18 at 9:08
1
$begingroup$
@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
$endgroup$
– user150203
Dec 27 '18 at 9:13