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While solving a problem, I got stuck at an integral. The integral is as follows:

Find the closed form of: $$I=\int _{ 0 }^{ 1 }{ \frac { \ln { x } { \left( \ln { \left( 1-{ x }^{ 2 } \right) } \right) }^{ 3 } }{ 1-x } dx } $$

I tried using power series but it failed. I tried various subtitutions which came to be of no use. Please help.

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    $\begingroup$ Why do you believe that this has a closed form solution? $\endgroup$ – Mark Viola Feb 26 '16 at 5:16
  • $\begingroup$ @Dr.MV I believe so because the question which I was solving had a good closed form. $\endgroup$ – Aditya Kumar Feb 26 '16 at 5:21
  • $\begingroup$ And how does that "question I was solving" relate to this integral? Just FYI ... one does not solve a question; one answers a question. ;-)) $\endgroup$ – Mark Viola Feb 26 '16 at 5:26
  • $\begingroup$ Thanks for correcting me @Dr.MV $\endgroup$ – Aditya Kumar Feb 26 '16 at 6:08
  • $\begingroup$ @MhenniBenghorbal I tried to use the power series of ln(1-x) and of 1/(1-x). It tried to get it to a form of harmonic numbers but I failed $\endgroup$ – Aditya Kumar Feb 26 '16 at 6:13
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We have $$ I=\int_{0}^{1}\frac{\log\left(x\right)\log^{3}\left(1-x^{2}\right)}{1-x}dx=\int_{0}^{1}\frac{\log\left(x\right)\log^{3}\left(1-x^{2}\right)}{1-x^{2}}dx+\int_{0}^{1}\frac{x\log\left(x\right)\log^{3}\left(1-x^{2}\right)}{1-x^{2}}dx $$ and so if we put $x=\sqrt{y}$ we get $$I=\frac{1}{4}\int_{0}^{1}\frac{y^{-1/2}\log\left(y\right)\log^{3}\left(1-y\right)}{1-y}dy+\frac{1}{4}\int_{0}^{1}\frac{\log\left(y\right)\log^{3}\left(1-y\right)}{1-y}dy$$ and recalling the definition of beta function $$B\left(a,b\right)=\int_{0}^{1}x^{a-1}\left(1-x\right)^{b-1}dx $$ we have $$\frac{\partial^{h+k}B}{\partial a^{h}\partial b^{k}}\left(a,b\right)=\int_{0}^{1}x^{a-1}\log^{h}\left(x\right)\left(1-x\right)^{b-1}\log^{k}\left(1-x\right)dx $$ hence $$I=\frac{1}{4}\frac{\partial^{4}B}{\partial a\partial b^{3}}\left(\frac{1}{2},0^{+}\right)+\frac{1}{4}\frac{\partial^{4}B}{\partial a\partial b^{3}}\left(1,0^{+}\right).$$ For the computation of the limit, we can use the asymptotic $\Gamma(x)=\frac{1}{x}+O(1)$ when $x\rightarrow 0$ and the relations between the polygamma terms and zeta.

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    $\begingroup$ Brilliant, since it can easily be generalized ! :-$)$ $\endgroup$ – Lucian Mar 1 '16 at 8:43
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The closed form is $$I=21\zeta({3})\log^22-\pi^2\log^32-\frac{13}{4}\pi^2\zeta({3})+48\zeta({5})-\frac{1}{4}\pi^4\log2$$

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    $\begingroup$ Can u post the steps also? $\endgroup$ – Aditya Kumar Feb 26 '16 at 12:14
  • $\begingroup$ Funny mathematica evaluates this to zero (using the standard "Integrate" routine, "NIntegrate" works) $\endgroup$ – tired Feb 26 '16 at 12:28
  • $\begingroup$ I based on my study of the integrals of the form $$\int_0^1\frac{\log{x}^p\log({1-x})^q\log{1+x})^r}{1±x}dx\ with\ p+q+r=4$$ Also $$\int _{ 0 }^{ 1 }{ \frac { \ln { x } { \left( \ln { \left( 1-{ x }^{ 2 } \right) } \right) }^{ 3 } }{ 1-x } dx }- \int _{ 0 }^{ 1 }{ \frac { \ln { x } { \left( \ln { \left( 1-{ x }^{ 2 } \right) } \right) }^{ 3 } }{ 1+x } dx } =3\zeta({5})$$ $\endgroup$ – user178256 Feb 26 '16 at 23:07
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    $\begingroup$ The two values coincide for the first $100$ decimals, which, given the reduced complexity of the expression, removes all reasonable doubt as to its veracity. $\endgroup$ – Lucian Feb 27 '16 at 0:55

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