Finding $limlimits_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$Evaluating $lim limits_nto infty,,, n!! intlimits_0^pi/2!! left(1-sqrt [n]sin x right),mathrm dx$Evaluate $int_0^1 frac1 x^2+2x+3,mathrm dx$Show how $fracpartialpartial x left[int_0^x (x-t)g(t),mathrmdtright] = int_0^x g(t),mathrmdt$Evaluate $large int_0^1left(frac1ln x + frac11-xright)^2 mathrm dx $ using elementary, high school techniquesFind the limit $lim_limitsxto 0^+left( e^frac1sin x-e^frac1xright)$L'Hopital rule to solve $ limlimits_xto 0 left(fracsin xxright)^frac1x^2 $Compute the limit $lim_ntoinfty nleft [int_0^frac pi4tan^n left ( fracxn right )mathrm dxright]^frac1n$How to find $limlimits_xto 0^+ frac1x int_0^x sin(frac1t),mathrm dt$$n^2 int_0^1 (1-x)^n sin(pi x) mathrmdx$If $I_n=int_0^1 fracx^nx^2+2019,mathrm dx$, evaluate $limlimits_nto infty nI_n$
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Finding $limlimits_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$
Evaluating $lim limits_nto infty,,, n!! intlimits_0^pi/2!! left(1-sqrt [n]sin x right),mathrm dx$Evaluate $int_0^1 frac1 x^2+2x+3,mathrm dx$Show how $fracpartialpartial x left[int_0^x (x-t)g(t),mathrmdtright] = int_0^x g(t),mathrmdt$Evaluate $large int_0^1left(frac1ln x + frac11-xright)^2 mathrm dx $ using elementary, high school techniquesFind the limit $lim_limitsxto 0^+left( e^frac1sin x-e^frac1xright)$L'Hopital rule to solve $ limlimits_xto 0 left(fracsin xxright)^frac1x^2 $Compute the limit $lim_ntoinfty nleft [int_0^frac pi4tan^n left ( fracxn right )mathrm dxright]^frac1n$How to find $limlimits_xto 0^+ frac1x int_0^x sin(frac1t),mathrm dt$$n^2 int_0^1 (1-x)^n sin(pi x) mathrmdx$If $I_n=int_0^1 fracx^nx^2+2019,mathrm dx$, evaluate $limlimits_nto infty nI_n$
$begingroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
edited 11 hours ago
rash
550115
asked 15 hours ago
赵家艺赵家艺
362
$endgroup$
|
show 1 more comment
$begingroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
edited 11 hours ago
rash
550115
asked 15 hours ago
赵家艺赵家艺
362
$endgroup$
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
15 hours ago
$begingroup$
there's some formula like $sum_ le N e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
15 hours ago
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
14 hours ago
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
14 hours ago
1
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
10 hours ago
|
show 1 more comment
$begingroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
edited 11 hours ago
rash
550115
asked 15 hours ago
赵家艺赵家艺
362
$endgroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
calculus integration limits
edited 11 hours ago
rash
550115
asked 15 hours ago
赵家艺赵家艺
362
edited 11 hours ago
rash
550115
asked 15 hours ago
赵家艺赵家艺
362
edited 11 hours ago
rash
550115
edited 11 hours ago
rash
550115
edited 11 hours ago
rash
550115
550115
asked 15 hours ago
赵家艺赵家艺
362
asked 15 hours ago
赵家艺赵家艺
362
asked 15 hours ago
赵家艺赵家艺
362
362
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
15 hours ago
$begingroup$
there's some formula like $sum_ le N e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
15 hours ago
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
14 hours ago
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
14 hours ago
1
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
10 hours ago
|
show 1 more comment
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
15 hours ago
$begingroup$
there's some formula like $sum_ le N e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
15 hours ago
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
14 hours ago
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
14 hours ago
1
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
10 hours ago
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
15 hours ago
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
15 hours ago
$begingroup$
there's some formula like $sum_ le N e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
15 hours ago
$begingroup$
there's some formula like $sum_ le N e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
15 hours ago
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
14 hours ago
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
14 hours ago
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
14 hours ago
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
14 hours ago
1
1
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
10 hours ago
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
10 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered 12 hours ago
TianlaluTianlalu
3,16721138
$endgroup$
add a comment |
$begingroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered 12 hours ago
TianlaluTianlalu
3,16721138
$endgroup$
add a comment |
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered 12 hours ago
TianlaluTianlalu
3,16721138
$endgroup$
add a comment |
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered 12 hours ago
TianlaluTianlalu
3,16721138
$endgroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered 12 hours ago
TianlaluTianlalu
3,16721138
answered 12 hours ago
TianlaluTianlalu
3,16721138
answered 12 hours ago
TianlaluTianlalu
3,16721138
answered 12 hours ago
TianlaluTianlalu
3,16721138
3,16721138
add a comment |
add a comment |
$begingroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
$endgroup$
add a comment |
$begingroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
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add a comment |
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Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
$endgroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
edited 11 hours ago
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
answered 13 hours ago
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,825620
3,825620
add a comment |
add a comment |
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(pinch theorem)
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– mathworker21
15 hours ago
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there's some formula like $sum_ le N e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
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– mathworker21
15 hours ago
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I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
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– Darkrai
14 hours ago
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After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
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– Dr. Wolfgang Hintze
14 hours ago
1
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Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
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– Sangchul Lee
10 hours ago