Mixing
Finding a closed form for $int_{0}^{1}frac{lnleft ( 1-x^{2} right )arcsin ^{2}x}{x^{2}}mathrm{d}xapprox...
$begingroup$
This question already has an answer here:
What is $int_0^1 frac{log left(1-x^2right) sin ^{-1}(x)^2}{x^2} , dx$?
1 answer
I'm attempting to find a closed form for
$$int_{0}^{1}frac{lnleft ( 1-x^{2} right )arcsin ^{2}x}{x^{2}}mathrm{d}xapprox -0.939332$$
I tried to use $$displaystyle arcsin^{2}x=frac{1}{2}sum_{k=1}^{infty }frac{left ( 2x right )^{2k}}{k^{2}dbinom{2k}{k}}$$
but it didn't work and became even more complicated. Any help would be appreciated.
calculus integration definite-integrals closed-form
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
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marked as duplicate by Carl Mummert, David H, Frank W., Community♦ Jan 16 at 4:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 1 more comment
$begingroup$
This question already has an answer here:
What is $int_0^1 frac{log left(1-x^2right) sin ^{-1}(x)^2}{x^2} , dx$?
1 answer
I'm attempting to find a closed form for
$$int_{0}^{1}frac{lnleft ( 1-x^{2} right )arcsin ^{2}x}{x^{2}}mathrm{d}xapprox -0.939332$$
I tried to use $$displaystyle arcsin^{2}x=frac{1}{2}sum_{k=1}^{infty }frac{left ( 2x right )^{2k}}{k^{2}dbinom{2k}{k}}$$
but it didn't work and became even more complicated. Any help would be appreciated.
calculus integration definite-integrals closed-form
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
$endgroup$
marked as duplicate by Carl Mummert, David H, Frank W., Community♦ Jan 16 at 4:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Have you tried letting $x=sintheta$ or some trigonometric substitution?
$endgroup$
– Frank W.
Jan 15 at 15:02
$begingroup$
@FrankW. yes,but failed
$endgroup$
– Renascence_5.
Jan 15 at 15:04
$begingroup$
@Renascence_5. Can we see your attempt using that substitution? Your work may be useful to us. By the way, [W|A] can't find one.
$endgroup$
– TheSimpliFire
Jan 15 at 15:08
$begingroup$
Maple also fails. So it seems likely that the indefinite integral is not elementary. There may remain a chance for the definite integral (but contour integration has been discouraged by the OP).
$endgroup$
– GEdgar
Jan 15 at 15:51
$begingroup$
@TheSimpliFire $-d left(frac{1}{x}-1right)$ integration by parts and use that substitution
$endgroup$
– Renascence_5.
Jan 15 at 16:01
|
show 1 more comment
$begingroup$
This question already has an answer here:
What is $int_0^1 frac{log left(1-x^2right) sin ^{-1}(x)^2}{x^2} , dx$?
1 answer
I'm attempting to find a closed form for
$$int_{0}^{1}frac{lnleft ( 1-x^{2} right )arcsin ^{2}x}{x^{2}}mathrm{d}xapprox -0.939332$$
I tried to use $$displaystyle arcsin^{2}x=frac{1}{2}sum_{k=1}^{infty }frac{left ( 2x right )^{2k}}{k^{2}dbinom{2k}{k}}$$
but it didn't work and became even more complicated. Any help would be appreciated.
calculus integration definite-integrals closed-form
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
$endgroup$
This question already has an answer here:
What is $int_0^1 frac{log left(1-x^2right) sin ^{-1}(x)^2}{x^2} , dx$?
1 answer
I'm attempting to find a closed form for
$$int_{0}^{1}frac{lnleft ( 1-x^{2} right )arcsin ^{2}x}{x^{2}}mathrm{d}xapprox -0.939332$$
I tried to use $$displaystyle arcsin^{2}x=frac{1}{2}sum_{k=1}^{infty }frac{left ( 2x right )^{2k}}{k^{2}dbinom{2k}{k}}$$
but it didn't work and became even more complicated. Any help would be appreciated.
This question already has an answer here:
What is $int_0^1 frac{log left(1-x^2right) sin ^{-1}(x)^2}{x^2} , dx$?
1 answer
calculus integration definite-integrals closed-form
calculus integration definite-integrals closed-form
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
edited Jan 15 at 23:49
Renascence_5.
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
asked Jan 15 at 15:00
Renascence_5.Renascence_5.
3,76812064
3,76812064
marked as duplicate by Carl Mummert, David H, Frank W., Community♦ Jan 16 at 4:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Carl Mummert, David H, Frank W., Community♦ Jan 16 at 4:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Have you tried letting $x=sintheta$ or some trigonometric substitution?
$endgroup$
– Frank W.
Jan 15 at 15:02
$begingroup$
@FrankW. yes,but failed
$endgroup$
– Renascence_5.
Jan 15 at 15:04
$begingroup$
@Renascence_5. Can we see your attempt using that substitution? Your work may be useful to us. By the way, [W|A] can't find one.
$endgroup$
– TheSimpliFire
Jan 15 at 15:08
$begingroup$
Maple also fails. So it seems likely that the indefinite integral is not elementary. There may remain a chance for the definite integral (but contour integration has been discouraged by the OP).
$endgroup$
– GEdgar
Jan 15 at 15:51
$begingroup$
@TheSimpliFire $-d left(frac{1}{x}-1right)$ integration by parts and use that substitution
$endgroup$
– Renascence_5.
Jan 15 at 16:01
|
show 1 more comment
$begingroup$
Have you tried letting $x=sintheta$ or some trigonometric substitution?
$endgroup$
– Frank W.
Jan 15 at 15:02
$begingroup$
@FrankW. yes,but failed
$endgroup$
– Renascence_5.
Jan 15 at 15:04
$begingroup$
@Renascence_5. Can we see your attempt using that substitution? Your work may be useful to us. By the way, [W|A] can't find one.
$endgroup$
– TheSimpliFire
Jan 15 at 15:08
$begingroup$
Maple also fails. So it seems likely that the indefinite integral is not elementary. There may remain a chance for the definite integral (but contour integration has been discouraged by the OP).
$endgroup$
– GEdgar
Jan 15 at 15:51
$begingroup$
@TheSimpliFire $-d left(frac{1}{x}-1right)$ integration by parts and use that substitution
$endgroup$
– Renascence_5.
Jan 15 at 16:01
$begingroup$
Have you tried letting $x=sintheta$ or some trigonometric substitution?
$endgroup$
– Frank W.
Jan 15 at 15:02
$begingroup$
Have you tried letting $x=sintheta$ or some trigonometric substitution?
$endgroup$
– Frank W.
Jan 15 at 15:02
$begingroup$
@FrankW. yes,but failed
$endgroup$
– Renascence_5.
Jan 15 at 15:04
$begingroup$
@FrankW. yes,but failed
$endgroup$
– Renascence_5.
Jan 15 at 15:04
$begingroup$
@Renascence_5. Can we see your attempt using that substitution? Your work may be useful to us. By the way, [W|A] can't find one.
$endgroup$
– TheSimpliFire
Jan 15 at 15:08
$begingroup$
@Renascence_5. Can we see your attempt using that substitution? Your work may be useful to us. By the way, [W|A] can't find one.
$endgroup$
– TheSimpliFire
Jan 15 at 15:08
$begingroup$
Maple also fails. So it seems likely that the indefinite integral is not elementary. There may remain a chance for the definite integral (but contour integration has been discouraged by the OP).
$endgroup$
– GEdgar
Jan 15 at 15:51
$begingroup$
Maple also fails. So it seems likely that the indefinite integral is not elementary. There may remain a chance for the definite integral (but contour integration has been discouraged by the OP).
$endgroup$
– GEdgar
Jan 15 at 15:51
$begingroup$
@TheSimpliFire $-d left(frac{1}{x}-1right)$ integration by parts and use that substitution
$endgroup$
– Renascence_5.
Jan 15 at 16:01
$begingroup$
@TheSimpliFire $-d left(frac{1}{x}-1right)$ integration by parts and use that substitution
$endgroup$
– Renascence_5.
Jan 15 at 16:01
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
The following closed-forms are proposed by Cornel Ioan Valean,
begin{equation*}
8 log (2)G+7 zeta (3)+frac{pi ^3}{4}- frac{pi ^2}{2} log (2)+pi log ^2(2)-16 Im( text{Li}_3(1+i))
end{equation*}
or
begin{equation*}
8 log (2)G+7 zeta (3)-frac{pi ^3}{2}- frac{pi ^2}{2} log (2)+4 , _4F_3left(frac{1}{2},frac{1}{2},1,1;frac{3}{2},frac{3}{2},frac{3}{2};1right).
end{equation*}
He used ideas from his upcoming book, (Almost) Impossible Integrals, Sums, and Series.
answered Jan 15 at 19:22
ZackyZacky
6,6701958
$endgroup$
$begingroup$
Thank you! But what method did he use. The book seems no ebook version.
$endgroup$
– Renascence_5.
Jan 16 at 0:14
$begingroup$
He only communicated me the closed-form which I shared with you. The book is not available now - it will appear in March, this year. If I get more details, I'll make an update later.
$endgroup$
– Zacky
Jan 16 at 1:06
1
$begingroup$
Now I know what the next present I'm gonna buy for myself is :)
$endgroup$
– clathratus
Jan 16 at 3:42
$begingroup$
Holy crap, I can't wait for this book!!
$endgroup$
– Frank W.
Jan 16 at 3:46
$begingroup$
@FrankW. me too!
$endgroup$
– Renascence_5.
Jan 16 at 4:08
|
show 1 more comment
$begingroup$
Using series, I get
$$ -2sum _{k=0}^{infty }{frac {{4}^{k} left( Psi left( k+3/2
right) +gamma right) left( k! right) ^{2}}{ left( 1+2,k
right) left( 2,k+2 right) !}}
$$
But I don't expect there to be a closed form.
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
A start
$$frac{arcsin(x)^2}{x^2}=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}x^{2n-2}$$
Thus
$$I=int_0^1frac{log(1-x^2)arcsin(x)^2}{x^2}mathrm dx=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Then we look at
$$J(n)=int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Wolfram Alpha produces
$$J(n)=frac1{1-2n}H_{n-1/2}$$
Where $H_n$ is the $n$-th harmonic number. This can also be written as
$$J(n)=frac1{1-2n}left[psi(n+1/2)+gammaright]$$
Where $psi(s)$ is the dilogarithm and $gamma$ is the Euler-Mascheroni constant.
That's all I've got for now.
answered Jan 16 at 4:41
clathratusclathratus
4,568337
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The following closed-forms are proposed by Cornel Ioan Valean,
begin{equation*}
8 log (2)G+7 zeta (3)+frac{pi ^3}{4}- frac{pi ^2}{2} log (2)+pi log ^2(2)-16 Im( text{Li}_3(1+i))
end{equation*}
or
begin{equation*}
8 log (2)G+7 zeta (3)-frac{pi ^3}{2}- frac{pi ^2}{2} log (2)+4 , _4F_3left(frac{1}{2},frac{1}{2},1,1;frac{3}{2},frac{3}{2},frac{3}{2};1right).
end{equation*}
He used ideas from his upcoming book, (Almost) Impossible Integrals, Sums, and Series.
answered Jan 15 at 19:22
ZackyZacky
6,6701958
$endgroup$
$begingroup$
Thank you! But what method did he use. The book seems no ebook version.
$endgroup$
– Renascence_5.
Jan 16 at 0:14
$begingroup$
He only communicated me the closed-form which I shared with you. The book is not available now - it will appear in March, this year. If I get more details, I'll make an update later.
$endgroup$
– Zacky
Jan 16 at 1:06
1
$begingroup$
Now I know what the next present I'm gonna buy for myself is :)
$endgroup$
– clathratus
Jan 16 at 3:42
$begingroup$
Holy crap, I can't wait for this book!!
$endgroup$
– Frank W.
Jan 16 at 3:46
$begingroup$
@FrankW. me too!
$endgroup$
– Renascence_5.
Jan 16 at 4:08
|
show 1 more comment
$begingroup$
The following closed-forms are proposed by Cornel Ioan Valean,
begin{equation*}
8 log (2)G+7 zeta (3)+frac{pi ^3}{4}- frac{pi ^2}{2} log (2)+pi log ^2(2)-16 Im( text{Li}_3(1+i))
end{equation*}
or
begin{equation*}
8 log (2)G+7 zeta (3)-frac{pi ^3}{2}- frac{pi ^2}{2} log (2)+4 , _4F_3left(frac{1}{2},frac{1}{2},1,1;frac{3}{2},frac{3}{2},frac{3}{2};1right).
end{equation*}
He used ideas from his upcoming book, (Almost) Impossible Integrals, Sums, and Series.
answered Jan 15 at 19:22
ZackyZacky
6,6701958
$endgroup$
$begingroup$
Thank you! But what method did he use. The book seems no ebook version.
$endgroup$
– Renascence_5.
Jan 16 at 0:14
$begingroup$
He only communicated me the closed-form which I shared with you. The book is not available now - it will appear in March, this year. If I get more details, I'll make an update later.
$endgroup$
– Zacky
Jan 16 at 1:06
1
$begingroup$
Now I know what the next present I'm gonna buy for myself is :)
$endgroup$
– clathratus
Jan 16 at 3:42
$begingroup$
Holy crap, I can't wait for this book!!
$endgroup$
– Frank W.
Jan 16 at 3:46
$begingroup$
@FrankW. me too!
$endgroup$
– Renascence_5.
Jan 16 at 4:08
|
show 1 more comment
$begingroup$
The following closed-forms are proposed by Cornel Ioan Valean,
begin{equation*}
8 log (2)G+7 zeta (3)+frac{pi ^3}{4}- frac{pi ^2}{2} log (2)+pi log ^2(2)-16 Im( text{Li}_3(1+i))
end{equation*}
or
begin{equation*}
8 log (2)G+7 zeta (3)-frac{pi ^3}{2}- frac{pi ^2}{2} log (2)+4 , _4F_3left(frac{1}{2},frac{1}{2},1,1;frac{3}{2},frac{3}{2},frac{3}{2};1right).
end{equation*}
He used ideas from his upcoming book, (Almost) Impossible Integrals, Sums, and Series.
answered Jan 15 at 19:22
ZackyZacky
6,6701958
$endgroup$
The following closed-forms are proposed by Cornel Ioan Valean,
begin{equation*}
8 log (2)G+7 zeta (3)+frac{pi ^3}{4}- frac{pi ^2}{2} log (2)+pi log ^2(2)-16 Im( text{Li}_3(1+i))
end{equation*}
or
begin{equation*}
8 log (2)G+7 zeta (3)-frac{pi ^3}{2}- frac{pi ^2}{2} log (2)+4 , _4F_3left(frac{1}{2},frac{1}{2},1,1;frac{3}{2},frac{3}{2},frac{3}{2};1right).
end{equation*}
He used ideas from his upcoming book, (Almost) Impossible Integrals, Sums, and Series.
answered Jan 15 at 19:22
ZackyZacky
6,6701958
answered Jan 15 at 19:22
ZackyZacky
6,6701958
answered Jan 15 at 19:22
ZackyZacky
6,6701958
answered Jan 15 at 19:22
ZackyZacky
6,6701958
6,6701958
$begingroup$
Thank you! But what method did he use. The book seems no ebook version.
$endgroup$
– Renascence_5.
Jan 16 at 0:14
$begingroup$
He only communicated me the closed-form which I shared with you. The book is not available now - it will appear in March, this year. If I get more details, I'll make an update later.
$endgroup$
– Zacky
Jan 16 at 1:06
1
$begingroup$
Now I know what the next present I'm gonna buy for myself is :)
$endgroup$
– clathratus
Jan 16 at 3:42
$begingroup$
Holy crap, I can't wait for this book!!
$endgroup$
– Frank W.
Jan 16 at 3:46
$begingroup$
@FrankW. me too!
$endgroup$
– Renascence_5.
Jan 16 at 4:08
|
show 1 more comment
$begingroup$
Thank you! But what method did he use. The book seems no ebook version.
$endgroup$
– Renascence_5.
Jan 16 at 0:14
$begingroup$
He only communicated me the closed-form which I shared with you. The book is not available now - it will appear in March, this year. If I get more details, I'll make an update later.
$endgroup$
– Zacky
Jan 16 at 1:06
1
$begingroup$
Now I know what the next present I'm gonna buy for myself is :)
$endgroup$
– clathratus
Jan 16 at 3:42
$begingroup$
Holy crap, I can't wait for this book!!
$endgroup$
– Frank W.
Jan 16 at 3:46
$begingroup$
@FrankW. me too!
$endgroup$
– Renascence_5.
Jan 16 at 4:08
$begingroup$
Thank you! But what method did he use. The book seems no ebook version.
$endgroup$
– Renascence_5.
Jan 16 at 0:14
$begingroup$
Thank you! But what method did he use. The book seems no ebook version.
$endgroup$
– Renascence_5.
Jan 16 at 0:14
$begingroup$
He only communicated me the closed-form which I shared with you. The book is not available now - it will appear in March, this year. If I get more details, I'll make an update later.
$endgroup$
– Zacky
Jan 16 at 1:06
$begingroup$
He only communicated me the closed-form which I shared with you. The book is not available now - it will appear in March, this year. If I get more details, I'll make an update later.
$endgroup$
– Zacky
Jan 16 at 1:06
1
1
$begingroup$
Now I know what the next present I'm gonna buy for myself is :)
$endgroup$
– clathratus
Jan 16 at 3:42
$begingroup$
Now I know what the next present I'm gonna buy for myself is :)
$endgroup$
– clathratus
Jan 16 at 3:42
$begingroup$
Holy crap, I can't wait for this book!!
$endgroup$
– Frank W.
Jan 16 at 3:46
$begingroup$
Holy crap, I can't wait for this book!!
$endgroup$
– Frank W.
Jan 16 at 3:46
$begingroup$
@FrankW. me too!
$endgroup$
– Renascence_5.
Jan 16 at 4:08
$begingroup$
@FrankW. me too!
$endgroup$
– Renascence_5.
Jan 16 at 4:08
|
show 1 more comment
$begingroup$
Using series, I get
$$ -2sum _{k=0}^{infty }{frac {{4}^{k} left( Psi left( k+3/2
right) +gamma right) left( k! right) ^{2}}{ left( 1+2,k
right) left( 2,k+2 right) !}}
$$
But I don't expect there to be a closed form.
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Using series, I get
$$ -2sum _{k=0}^{infty }{frac {{4}^{k} left( Psi left( k+3/2
right) +gamma right) left( k! right) ^{2}}{ left( 1+2,k
right) left( 2,k+2 right) !}}
$$
But I don't expect there to be a closed form.
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Using series, I get
$$ -2sum _{k=0}^{infty }{frac {{4}^{k} left( Psi left( k+3/2
right) +gamma right) left( k! right) ^{2}}{ left( 1+2,k
right) left( 2,k+2 right) !}}
$$
But I don't expect there to be a closed form.
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
$endgroup$
Using series, I get
$$ -2sum _{k=0}^{infty }{frac {{4}^{k} left( Psi left( k+3/2
right) +gamma right) left( k! right) ^{2}}{ left( 1+2,k
right) left( 2,k+2 right) !}}
$$
But I don't expect there to be a closed form.
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 16:21
Robert IsraelRobert Israel
324k23214468
324k23214468
add a comment |
add a comment |
$begingroup$
A start
$$frac{arcsin(x)^2}{x^2}=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}x^{2n-2}$$
Thus
$$I=int_0^1frac{log(1-x^2)arcsin(x)^2}{x^2}mathrm dx=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Then we look at
$$J(n)=int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Wolfram Alpha produces
$$J(n)=frac1{1-2n}H_{n-1/2}$$
Where $H_n$ is the $n$-th harmonic number. This can also be written as
$$J(n)=frac1{1-2n}left[psi(n+1/2)+gammaright]$$
Where $psi(s)$ is the dilogarithm and $gamma$ is the Euler-Mascheroni constant.
That's all I've got for now.
answered Jan 16 at 4:41
clathratusclathratus
4,568337
$endgroup$
add a comment |
$begingroup$
A start
$$frac{arcsin(x)^2}{x^2}=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}x^{2n-2}$$
Thus
$$I=int_0^1frac{log(1-x^2)arcsin(x)^2}{x^2}mathrm dx=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Then we look at
$$J(n)=int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Wolfram Alpha produces
$$J(n)=frac1{1-2n}H_{n-1/2}$$
Where $H_n$ is the $n$-th harmonic number. This can also be written as
$$J(n)=frac1{1-2n}left[psi(n+1/2)+gammaright]$$
Where $psi(s)$ is the dilogarithm and $gamma$ is the Euler-Mascheroni constant.
That's all I've got for now.
answered Jan 16 at 4:41
clathratusclathratus
4,568337
$endgroup$
add a comment |
$begingroup$
A start
$$frac{arcsin(x)^2}{x^2}=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}x^{2n-2}$$
Thus
$$I=int_0^1frac{log(1-x^2)arcsin(x)^2}{x^2}mathrm dx=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Then we look at
$$J(n)=int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Wolfram Alpha produces
$$J(n)=frac1{1-2n}H_{n-1/2}$$
Where $H_n$ is the $n$-th harmonic number. This can also be written as
$$J(n)=frac1{1-2n}left[psi(n+1/2)+gammaright]$$
Where $psi(s)$ is the dilogarithm and $gamma$ is the Euler-Mascheroni constant.
That's all I've got for now.
answered Jan 16 at 4:41
clathratusclathratus
4,568337
$endgroup$
A start
$$frac{arcsin(x)^2}{x^2}=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}x^{2n-2}$$
Thus
$$I=int_0^1frac{log(1-x^2)arcsin(x)^2}{x^2}mathrm dx=sum_{ngeq1}frac{2^{2n-1}}{n^2{2nchoose n}}int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Then we look at
$$J(n)=int_0^1x^{2n-2}log(1-x^2)mathrm dx$$
Wolfram Alpha produces
$$J(n)=frac1{1-2n}H_{n-1/2}$$
Where $H_n$ is the $n$-th harmonic number. This can also be written as
$$J(n)=frac1{1-2n}left[psi(n+1/2)+gammaright]$$
Where $psi(s)$ is the dilogarithm and $gamma$ is the Euler-Mascheroni constant.
That's all I've got for now.
answered Jan 16 at 4:41
clathratusclathratus
4,568337
answered Jan 16 at 4:41
clathratusclathratus
4,568337
answered Jan 16 at 4:41
clathratusclathratus
4,568337
answered Jan 16 at 4:41
clathratusclathratus
4,568337
4,568337
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Have you tried letting $x=sintheta$ or some trigonometric substitution?
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– Frank W.
Jan 15 at 15:02
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@FrankW. yes,but failed
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– Renascence_5.
Jan 15 at 15:04
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@Renascence_5. Can we see your attempt using that substitution? Your work may be useful to us. By the way, [W|A] can't find one.
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– TheSimpliFire
Jan 15 at 15:08
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Maple also fails. So it seems likely that the indefinite integral is not elementary. There may remain a chance for the definite integral (but contour integration has been discouraged by the OP).
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– GEdgar
Jan 15 at 15:51
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@TheSimpliFire $-d left(frac{1}{x}-1right)$ integration by parts and use that substitution
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– Renascence_5.
Jan 15 at 16:01