# A closed form for the dilogarithm integral $\int _{ 0 }^{ 1 }{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx }$

$$\int _{ 0 }^{ 1 }{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx }$$ when I was solving an infinite series by using the beta function I encountered the above dilogarithm integral which has a quadratic term. I tried to apply u-substitution and other dilogarithm identities to evaluate it, Unfortunately, I didn't get it. Since the infinite series has a closed-form in terms of Apery$$\zeta \left( 3 \right)$$ and Catalan constant $$G$$, The result of this integral would be to in such constants. I hope if anyone could come up with a good method to evaluate this integral.

• By the way, Welcome to the site ! – Claude Leibovici Jun 4 '20 at 8:51
• I think that I made it. Have a look at my edit. – Claude Leibovici Jun 5 '20 at 3:52

We know that $$\text{Li}_2(x)=-\int_0^1 \frac{x\ln(y)}{1-xy}dy$$

so

$$\int_0^1 \frac{\text{Li}_2(2x(1-x))}{x}dx=-2\int_0^1 \ln(y)\left[\int_0^1 \frac{1-x}{1-2x(1-x)y}dx\right]dy$$

$$=-2\int_0^1 \frac{\ln(y)\arctan\sqrt{\frac{y}{2-y}}}{\sqrt{y(2-y)}}dy\overset{\sqrt{\frac{y}{2-y}}=x}{=}-4\int_0^1\frac{\ln\left(\frac{2x^2}{1+x^2}\right)\arctan(x)}{1+x^2}dx$$

$$\overset{x=\tan\theta}{=}-4\int_0^{\pi/4}\theta\ln(2\sin^2 \theta)\ d\theta=-4\ln(2)\int_0^{\pi/4}\theta\ d\theta-8\int_0^{\pi/4}\theta\ln(\sin\theta)\ d\theta$$

$$=-4\ln(2)\left(\frac{\pi^2}{32}\right)-8\left(\frac{35}{128}\zeta(3)-\frac{\pi^2}{32}\ln(2)-\frac{\pi}{8}G\right)$$

$$=\pi G-\frac{35}{16}\zeta(3)+\frac{\pi^2}{8}\ln(2).$$

The last integral follows from using the Fourier series of $$\ln(\sin\theta)=-\ln(2)-\sum_{n=1}^\infty\frac{\cos(2n\theta)}{n}$$.

• What a beautiful solution sir, you made my day. I think That log integral for dilogarithm series comes in handy while solving such dilog/polylog integrals – Syed Jun 7 '20 at 7:05
• Thank you . Glad you found it helpful :). Yes that log representation is very useful and can be found generalized in the book ( almost impossible integrals , sums and series) page 3 or 4 i think. – Ali Shadhar Jun 7 '20 at 7:46

Mathematica gives $$\pi C+\frac{1}{16} \left(\pi ^2 \log (4)-35 \zeta (3)\right).$$

• thanks for the closed-form, but it would be better if you could show me how to evaluate this integral – Syed Jun 3 '20 at 19:03

Considering $$I(x)=\int{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx }$$ please find (and enjoy) the antiderivative.

Using it $$I(1)=-2 \zeta (3)-\frac{1}{6} \pi ^2 \log (2)$$ $$I(0)=-\pi C+\frac{3 \zeta (3)}{16}-\frac{7}{24} \pi ^2 \log (2)$$

$$J=\int_0^1{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx }=\pi C-\frac{35}{16} \zeta (3)+\frac{1}{8} \pi ^2 \log (2)$$

Edit

Consider that $$\text{Li}_2(t)=\sum_{n=1}^\infty \frac t {n^2}\implies \text{Li}_2(2x(1-x))=\sum_{n=1}^\infty \frac {\big[2x(1-x)\big]^n} {n^2}$$

$$\int{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx }=\sum_{n=1}^\infty \frac {2^n}{n^2} \int (1-x)^n x^{n-1}\,dx$$ $$J=\int_0^1{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx }=\sum_{n=1}^\infty\frac{2^n}{n^2}\frac{\Gamma (n) \,\,\Gamma (n+1)}{\Gamma (2 n+1)}=\sqrt{\pi }\sum_{n=1}^\infty\frac{ 2^{-n}\,\, \Gamma (n)}{n^2 \,\Gamma \left(n+\frac{1}{2}\right)}$$

$$J=\, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;\frac{1}{2}\right)=\pi C-\frac{35 \zeta (3)}{16}+\frac{1}{8} \pi ^2 \log (2)$$

• Although the anti derivative seems to be quite daunting, Thanks for your valuable information. – Syed Jun 4 '20 at 8:39
• @Syed. It is not daunting. It is a monster ! – Claude Leibovici Jun 4 '20 at 8:50
• Yes your right thats a monster, But what if there could be a way to tame this monster by making use of some advance techniques without integrating it directly. :) – Syed Jun 4 '20 at 9:04