Tricky Integral — $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$Probability that three independent uniform $(0,1)$ random variables can form a triangleDefinite Integral $int_0^rarcsinleft(fracsqrtr^2-x^2R^2right)dx$Integral $int_0^infty frac1(alpha x^2 + 1) left(- 2 sqrtfrac x^2x^2+1+2 x+pi right) , dx$Closed Form for $int_0^1 fraclog(x)sqrt1-x^2sqrtx^2+2+2sqrt2dx$Integral $int_0^1 log frac1+ax1-axfracdxxsqrt1-x^2=piarcsin a$Closed-form of integral $int_0^1 int_0^1 fracarcsinleft(sqrt1-ssqrtyright)sqrt1-y cdot (sy-y+1),ds,dy $An integral by O. Furdui $int_0^1 log^2(sqrt1+x-sqrt1-x) dx$How to solve this definite integral $int_0^1fracat+bsqrtct^2+dt+farctan(fracgt+hsqrtct^2+dt+f)dt$Integration giving $arcsin x$ by a unique method: $sin^-1t=int_0^tfracdxsqrt1-x^2$Integral $int_0^fracpi2 arcsin(sqrtsin x) dx$Calculating improper integral $intlimits_0^1 fracarcsin(x)sqrt1-x^2,dx$
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Tricky Integral — $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$
Probability that three independent uniform $(0,1)$ random variables can form a triangleDefinite Integral $int_0^rarcsinleft(fracsqrtr^2-x^2R^2right)dx$Integral $int_0^infty frac1(alpha x^2 + 1) left(- 2 sqrtfrac x^2x^2+1+2 x+pi right) , dx$Closed Form for $int_0^1 fraclog(x)sqrt1-x^2sqrtx^2+2+2sqrt2dx$Integral $int_0^1 log frac1+ax1-axfracdxxsqrt1-x^2=piarcsin a$Closed-form of integral $int_0^1 int_0^1 fracarcsinleft(sqrt1-ssqrtyright)sqrt1-y cdot (sy-y+1),ds,dy $An integral by O. Furdui $int_0^1 log^2(sqrt1+x-sqrt1-x) dx$How to solve this definite integral $int_0^1fracat+bsqrtct^2+dt+farctan(fracgt+hsqrtct^2+dt+f)dt$Integration giving $arcsin x$ by a unique method: $sin^-1t=int_0^tfracdxsqrt1-x^2$Integral $int_0^fracpi2 arcsin(sqrtsin x) dx$Calculating improper integral $intlimits_0^1 fracarcsin(x)sqrt1-x^2,dx$
$begingroup$
TL;DR: I can't get a closed form for the integral below.
$$ int_0^1 sqrtx^2-4x+3 arcsin(x)~dx $$
I got an interesting question from a coworker a while ago:
Question:
The quantities $a$, $b$, and $c$ are chosen uniformly and independently from $[0, 1]$.
a) What is the probability a triangle can be constructed with $a$, $b$, and $c$ as side lengths?
b) Given we can form such a triangle, what is its expected area?
I can do a) pretty easily -- each constraint like $a < b + c$ cuts off a corner of the cube with area $1/6$, and the cut-off bits are disjoint, so the remaining area is $1/2$.
Part b) is where things get hairy. I can reduce the problem down to a single integral. I feel like it should be expressible in terms of known constants, though I admit I have no good reason to believe this.
$$
frac340 int_0^1 x sqrt3-4x+x^2 left( sqrt1 - x^2 + fracarcsinxx right)~dx
$$
That can be split into two parts:
$$
frac340 int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx + frac340 int_0^1 sqrt3-4x+x^2 arcsin(x)~dx
$$
The first part can be solved exactly.
$$
beginalign*
int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx &= int_0^1 x sqrt(3-x)(1-x)(1-x)(1+x)~dx \
&= int_0^1 x(1-x) sqrt(3-x)(1+x)~dx \
&= frac112 (32 - 9 sqrt3 - 4pi) textrm by Mathematica
endalign*
$$
The second part is still pretty stubborn.
Mathematica tells me the integral (without the 3/40 constant) is approximately 0.452854, but doesn't given an exact form. Does anyone have any ideas how to evaluate this further?
integration definite-integrals
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
$endgroup$
|
show 8 more comments
$begingroup$
TL;DR: I can't get a closed form for the integral below.
$$ int_0^1 sqrtx^2-4x+3 arcsin(x)~dx $$
I got an interesting question from a coworker a while ago:
Question:
The quantities $a$, $b$, and $c$ are chosen uniformly and independently from $[0, 1]$.
a) What is the probability a triangle can be constructed with $a$, $b$, and $c$ as side lengths?
b) Given we can form such a triangle, what is its expected area?
I can do a) pretty easily -- each constraint like $a < b + c$ cuts off a corner of the cube with area $1/6$, and the cut-off bits are disjoint, so the remaining area is $1/2$.
Part b) is where things get hairy. I can reduce the problem down to a single integral. I feel like it should be expressible in terms of known constants, though I admit I have no good reason to believe this.
$$
frac340 int_0^1 x sqrt3-4x+x^2 left( sqrt1 - x^2 + fracarcsinxx right)~dx
$$
That can be split into two parts:
$$
frac340 int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx + frac340 int_0^1 sqrt3-4x+x^2 arcsin(x)~dx
$$
The first part can be solved exactly.
$$
beginalign*
int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx &= int_0^1 x sqrt(3-x)(1-x)(1-x)(1+x)~dx \
&= int_0^1 x(1-x) sqrt(3-x)(1+x)~dx \
&= frac112 (32 - 9 sqrt3 - 4pi) textrm by Mathematica
endalign*
$$
The second part is still pretty stubborn.
Mathematica tells me the integral (without the 3/40 constant) is approximately 0.452854, but doesn't given an exact form. Does anyone have any ideas how to evaluate this further?
integration definite-integrals
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
$endgroup$
$begingroup$
math.stackexchange.com/questions/253893/…
$endgroup$
– John Wayland Bales
Mar 24 at 21:56
1
$begingroup$
@GReyes This was obviously already done. You can check the expression for the final integral.
$endgroup$
– user
Mar 24 at 22:27
1
$begingroup$
So what is the question? The textpaste link doesn't work, by the way ...
$endgroup$
– Math-fun
2 days ago
1
$begingroup$
@Math-fun I guess to evaluate the, most likely, non-elementary integral. The link works at least for me.
$endgroup$
– mrtaurho
2 days ago
4
$begingroup$
With the help of Mathematica I found that $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$ conjecturally evaluates to $$-2 G+frac12 i left(textLi_2left(frac-i+sqrt3sqrt3+1right)-textLi_2left(fraci+sqrt3sqrt3+1right)+textLi_2left(frac1-i(2-i)+sqrt3right)-textLi_2left(frac1+i(2+i)+sqrt3right)right)-frac7 sqrt34+fracpi 3+3-frac112 pi log left(14-8 sqrt3right),$$ where G is Catalan's Constant. The hope is that this ugly expression can be simplified further.
$endgroup$
– James Arathoon
2 days ago
|
show 8 more comments
$begingroup$
TL;DR: I can't get a closed form for the integral below.
$$ int_0^1 sqrtx^2-4x+3 arcsin(x)~dx $$
I got an interesting question from a coworker a while ago:
Question:
The quantities $a$, $b$, and $c$ are chosen uniformly and independently from $[0, 1]$.
a) What is the probability a triangle can be constructed with $a$, $b$, and $c$ as side lengths?
b) Given we can form such a triangle, what is its expected area?
I can do a) pretty easily -- each constraint like $a < b + c$ cuts off a corner of the cube with area $1/6$, and the cut-off bits are disjoint, so the remaining area is $1/2$.
Part b) is where things get hairy. I can reduce the problem down to a single integral. I feel like it should be expressible in terms of known constants, though I admit I have no good reason to believe this.
$$
frac340 int_0^1 x sqrt3-4x+x^2 left( sqrt1 - x^2 + fracarcsinxx right)~dx
$$
That can be split into two parts:
$$
frac340 int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx + frac340 int_0^1 sqrt3-4x+x^2 arcsin(x)~dx
$$
The first part can be solved exactly.
$$
beginalign*
int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx &= int_0^1 x sqrt(3-x)(1-x)(1-x)(1+x)~dx \
&= int_0^1 x(1-x) sqrt(3-x)(1+x)~dx \
&= frac112 (32 - 9 sqrt3 - 4pi) textrm by Mathematica
endalign*
$$
The second part is still pretty stubborn.
Mathematica tells me the integral (without the 3/40 constant) is approximately 0.452854, but doesn't given an exact form. Does anyone have any ideas how to evaluate this further?
integration definite-integrals
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
$endgroup$
TL;DR: I can't get a closed form for the integral below.
$$ int_0^1 sqrtx^2-4x+3 arcsin(x)~dx $$
I got an interesting question from a coworker a while ago:
Question:
The quantities $a$, $b$, and $c$ are chosen uniformly and independently from $[0, 1]$.
a) What is the probability a triangle can be constructed with $a$, $b$, and $c$ as side lengths?
b) Given we can form such a triangle, what is its expected area?
I can do a) pretty easily -- each constraint like $a < b + c$ cuts off a corner of the cube with area $1/6$, and the cut-off bits are disjoint, so the remaining area is $1/2$.
Part b) is where things get hairy. I can reduce the problem down to a single integral. I feel like it should be expressible in terms of known constants, though I admit I have no good reason to believe this.
$$
frac340 int_0^1 x sqrt3-4x+x^2 left( sqrt1 - x^2 + fracarcsinxx right)~dx
$$
That can be split into two parts:
$$
frac340 int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx + frac340 int_0^1 sqrt3-4x+x^2 arcsin(x)~dx
$$
The first part can be solved exactly.
$$
beginalign*
int_0^1 x sqrt(3-4x+x^2)(1 - x^2)~dx &= int_0^1 x sqrt(3-x)(1-x)(1-x)(1+x)~dx \
&= int_0^1 x(1-x) sqrt(3-x)(1+x)~dx \
&= frac112 (32 - 9 sqrt3 - 4pi) textrm by Mathematica
endalign*
$$
The second part is still pretty stubborn.
Mathematica tells me the integral (without the 3/40 constant) is approximately 0.452854, but doesn't given an exact form. Does anyone have any ideas how to evaluate this further?
integration definite-integrals
integration definite-integrals
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
edited Mar 25 at 4:31
Henry Swanson
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
asked Mar 24 at 21:46
Henry SwansonHenry Swanson
10.1k12355
10.1k12355
$begingroup$
math.stackexchange.com/questions/253893/…
$endgroup$
– John Wayland Bales
Mar 24 at 21:56
1
$begingroup$
@GReyes This was obviously already done. You can check the expression for the final integral.
$endgroup$
– user
Mar 24 at 22:27
1
$begingroup$
So what is the question? The textpaste link doesn't work, by the way ...
$endgroup$
– Math-fun
2 days ago
1
$begingroup$
@Math-fun I guess to evaluate the, most likely, non-elementary integral. The link works at least for me.
$endgroup$
– mrtaurho
2 days ago
4
$begingroup$
With the help of Mathematica I found that $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$ conjecturally evaluates to $$-2 G+frac12 i left(textLi_2left(frac-i+sqrt3sqrt3+1right)-textLi_2left(fraci+sqrt3sqrt3+1right)+textLi_2left(frac1-i(2-i)+sqrt3right)-textLi_2left(frac1+i(2+i)+sqrt3right)right)-frac7 sqrt34+fracpi 3+3-frac112 pi log left(14-8 sqrt3right),$$ where G is Catalan's Constant. The hope is that this ugly expression can be simplified further.
$endgroup$
– James Arathoon
2 days ago
|
show 8 more comments
$begingroup$
math.stackexchange.com/questions/253893/…
$endgroup$
– John Wayland Bales
Mar 24 at 21:56
1
$begingroup$
@GReyes This was obviously already done. You can check the expression for the final integral.
$endgroup$
– user
Mar 24 at 22:27
1
$begingroup$
So what is the question? The textpaste link doesn't work, by the way ...
$endgroup$
– Math-fun
2 days ago
1
$begingroup$
@Math-fun I guess to evaluate the, most likely, non-elementary integral. The link works at least for me.
$endgroup$
– mrtaurho
2 days ago
4
$begingroup$
With the help of Mathematica I found that $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$ conjecturally evaluates to $$-2 G+frac12 i left(textLi_2left(frac-i+sqrt3sqrt3+1right)-textLi_2left(fraci+sqrt3sqrt3+1right)+textLi_2left(frac1-i(2-i)+sqrt3right)-textLi_2left(frac1+i(2+i)+sqrt3right)right)-frac7 sqrt34+fracpi 3+3-frac112 pi log left(14-8 sqrt3right),$$ where G is Catalan's Constant. The hope is that this ugly expression can be simplified further.
$endgroup$
– James Arathoon
2 days ago
$begingroup$
math.stackexchange.com/questions/253893/…
$endgroup$
– John Wayland Bales
Mar 24 at 21:56
$begingroup$
math.stackexchange.com/questions/253893/…
$endgroup$
– John Wayland Bales
Mar 24 at 21:56
1
1
$begingroup$
@GReyes This was obviously already done. You can check the expression for the final integral.
$endgroup$
– user
Mar 24 at 22:27
$begingroup$
@GReyes This was obviously already done. You can check the expression for the final integral.
$endgroup$
– user
Mar 24 at 22:27
1
1
$begingroup$
So what is the question? The textpaste link doesn't work, by the way ...
$endgroup$
– Math-fun
2 days ago
$begingroup$
So what is the question? The textpaste link doesn't work, by the way ...
$endgroup$
– Math-fun
2 days ago
1
1
$begingroup$
@Math-fun I guess to evaluate the, most likely, non-elementary integral. The link works at least for me.
$endgroup$
– mrtaurho
2 days ago
$begingroup$
@Math-fun I guess to evaluate the, most likely, non-elementary integral. The link works at least for me.
$endgroup$
– mrtaurho
2 days ago
4
4
$begingroup$
With the help of Mathematica I found that $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$ conjecturally evaluates to $$-2 G+frac12 i left(textLi_2left(frac-i+sqrt3sqrt3+1right)-textLi_2left(fraci+sqrt3sqrt3+1right)+textLi_2left(frac1-i(2-i)+sqrt3right)-textLi_2left(frac1+i(2+i)+sqrt3right)right)-frac7 sqrt34+fracpi 3+3-frac112 pi log left(14-8 sqrt3right),$$ where G is Catalan's Constant. The hope is that this ugly expression can be simplified further.
$endgroup$
– James Arathoon
2 days ago
$begingroup$
With the help of Mathematica I found that $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$ conjecturally evaluates to $$-2 G+frac12 i left(textLi_2left(frac-i+sqrt3sqrt3+1right)-textLi_2left(fraci+sqrt3sqrt3+1right)+textLi_2left(frac1-i(2-i)+sqrt3right)-textLi_2left(frac1+i(2+i)+sqrt3right)right)-frac7 sqrt34+fracpi 3+3-frac112 pi log left(14-8 sqrt3right),$$ where G is Catalan's Constant. The hope is that this ugly expression can be simplified further.
$endgroup$
– James Arathoon
2 days ago
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This is not an answer.
We could use
$$sqrtx^2-4x+3=sum_n=0^infty a_n, x^n$$ with
$$a_n=frac2(2 n-3), a_n-1-(n-3), a_n-23 n qquad textwhereqquad a_0=sqrt3qquad a_1=-frac2sqrt3$$ and
$$int_0^1 x^narcsin(x),dx=fracpi 2( n+1)-fracsqrtpi ,,Gamma left(fracn2+1right)(n+1)^2
,, Gamma left(fracn+12right)$$ but the convergence is very slow.
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
Your Answer
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1 Answer
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$begingroup$
This is not an answer.
We could use
$$sqrtx^2-4x+3=sum_n=0^infty a_n, x^n$$ with
$$a_n=frac2(2 n-3), a_n-1-(n-3), a_n-23 n qquad textwhereqquad a_0=sqrt3qquad a_1=-frac2sqrt3$$ and
$$int_0^1 x^narcsin(x),dx=fracpi 2( n+1)-fracsqrtpi ,,Gamma left(fracn2+1right)(n+1)^2
,, Gamma left(fracn+12right)$$ but the convergence is very slow.
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
This is not an answer.
We could use
$$sqrtx^2-4x+3=sum_n=0^infty a_n, x^n$$ with
$$a_n=frac2(2 n-3), a_n-1-(n-3), a_n-23 n qquad textwhereqquad a_0=sqrt3qquad a_1=-frac2sqrt3$$ and
$$int_0^1 x^narcsin(x),dx=fracpi 2( n+1)-fracsqrtpi ,,Gamma left(fracn2+1right)(n+1)^2
,, Gamma left(fracn+12right)$$ but the convergence is very slow.
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
This is not an answer.
We could use
$$sqrtx^2-4x+3=sum_n=0^infty a_n, x^n$$ with
$$a_n=frac2(2 n-3), a_n-1-(n-3), a_n-23 n qquad textwhereqquad a_0=sqrt3qquad a_1=-frac2sqrt3$$ and
$$int_0^1 x^narcsin(x),dx=fracpi 2( n+1)-fracsqrtpi ,,Gamma left(fracn2+1right)(n+1)^2
,, Gamma left(fracn+12right)$$ but the convergence is very slow.
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
This is not an answer.
We could use
$$sqrtx^2-4x+3=sum_n=0^infty a_n, x^n$$ with
$$a_n=frac2(2 n-3), a_n-1-(n-3), a_n-23 n qquad textwhereqquad a_0=sqrt3qquad a_1=-frac2sqrt3$$ and
$$int_0^1 x^narcsin(x),dx=fracpi 2( n+1)-fracsqrtpi ,,Gamma left(fracn2+1right)(n+1)^2
,, Gamma left(fracn+12right)$$ but the convergence is very slow.
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 22 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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$begingroup$
math.stackexchange.com/questions/253893/…
$endgroup$
– John Wayland Bales
Mar 24 at 21:56
1
$begingroup$
@GReyes This was obviously already done. You can check the expression for the final integral.
$endgroup$
– user
Mar 24 at 22:27
1
$begingroup$
So what is the question? The textpaste link doesn't work, by the way ...
$endgroup$
– Math-fun
2 days ago
1
$begingroup$
@Math-fun I guess to evaluate the, most likely, non-elementary integral. The link works at least for me.
$endgroup$
– mrtaurho
2 days ago
4
$begingroup$
With the help of Mathematica I found that $int_0^1 sqrtx^2-4x+3 arcsin(x)~dx$ conjecturally evaluates to $$-2 G+frac12 i left(textLi_2left(frac-i+sqrt3sqrt3+1right)-textLi_2left(fraci+sqrt3sqrt3+1right)+textLi_2left(frac1-i(2-i)+sqrt3right)-textLi_2left(frac1+i(2+i)+sqrt3right)right)-frac7 sqrt34+fracpi 3+3-frac112 pi log left(14-8 sqrt3right),$$ where G is Catalan's Constant. The hope is that this ugly expression can be simplified further.
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– James Arathoon
2 days ago