By using dilogarithm functional equations we can show that $$ \int_0^1 \operatorname{Li}_2\left(1-x^2\right)\,dx = \frac{\pi^2}{2}-4, $$ where $\operatorname{Li}_2$ is the dilogarithm function.

Could we evaluate in closed-form the following integral?

$$ I = \int_0^1 \operatorname{Li}_3\left(1-x^2\right)\,dx, $$

where $\operatorname{Li}_3$ is the trilogarithm function.

A related integral with known closed-form is $$\int_0^1 \operatorname{Li}_3\left(\frac{1}{x^2}\right)\,dx = \zeta(3)+\frac{\pi^2}{3}-8\ln2 - 4\pi\,i,$$ where $\zeta$ is the Riemann zeta function.

  • $\begingroup$ I doubt that the integral admits a closed form, because there exists a functional equation to simplify $Li_2(1-x^2)$ but this isn't the case for the trilogarithm. $\endgroup$ – Redundant Aunt Aug 21 '15 at 21:44
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    $\begingroup$ another approach would be to expand the integral to an infinite series $$I=\sum_{n=1}^{\infty} \frac{2^{2n}}{n^3(2n+1)\binom{2n}{n}}$$ and then calculate definite integrals of $\arcsin^2 x$ ,which has similar series representation $\endgroup$ – nospoon Aug 21 '15 at 23:12

$$-\frac72\zeta\left(3\right)+\pi^2\left(\ln 2-1\right)+8$$

  • $\begingroup$ Nice but how did you arrived here sir ? $\endgroup$ – Renato Faraone Aug 21 '15 at 22:43
  • $\begingroup$ @RenatoFaraone Namagiri was actually looking for Kleo, but found only L.G. $\endgroup$ – Start wearing purple Aug 21 '15 at 22:48
  • $\begingroup$ Probably it was repeated integration by parts. By this way the problem boils down to integrate the product of $\left(-x-\frac{1}{2}\log(1-x)+\frac{1}{2}\log(1+x)\right)$ and $\frac{x\log x}{1-x^2}$. $\endgroup$ – Jack D'Aurizio Aug 21 '15 at 22:56
  • $\begingroup$ Using differentiation under the integral sign, that boils down to evaluating a few derivatives of a Beta function, not that hard. $\endgroup$ – Jack D'Aurizio Aug 21 '15 at 22:59
  • $\begingroup$ @JackD'Aurizio Feel free to post your solution. Thank you L.G. for your answer. +1. $\endgroup$ – user153012 Aug 21 '15 at 23:00

Expanding my comment: the substitution $1-x^2\mapsto x$, followed by expanding the trilogarithm and keeping in mind Legrende's duplication formula $B(n+1,\frac12)=2^{2n+1}B(n+1,n+1)$, we arrive at $$I=\sum_{n=1}^{\infty} \frac{2^{2n}}{n^3(2n+1)\binom{2n}{n}}=\sum_{n=1}^{\infty}\frac{2^{2n}}{n^3\binom{2n}{n}}-\sum_{n=1}^{\infty}\frac{2^{2n+1}}{n^2(2n+1)\binom{2n}{n}}\\=4\int_0^1\frac{\arcsin^2x}{x}\,dx-4\int_0^1\arcsin^2x \,dx$$ where I used the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{(2x)^{2n}}{n^2\binom{2n}{n}}=2\arcsin^2x$.

The second integral is easily evaluated by IBP twice: $$\int_0^1\arcsin^2x \,dx=\int_0^{\frac{\pi}{2}}x^2\cos x \,dx=\frac{\pi^2}{4}-2$$

The first integral may be evaluated by IBP and using that $\displaystyle -\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}$: $$\begin{align} \int_0^1\frac{\arcsin^2x}{x}\,dx\\&=\int_0^{\frac{\pi}{2}}x^2\cot x \,dx\\&=-2\int_0^{\frac{\pi}{2}}x\ln\sin x \,dx\\ &=2\int_0^{\frac{\pi}{2}}x\left(\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}\right)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\int_0^{\frac{\pi}{2}}x\cos(2nx)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\frac1{4n^2}((-1)^n-1)\\&=\frac{\pi^2}{4}\ln2-\frac{7}{8}\zeta(3). \end{align}$$

  • $\begingroup$ Well Done! A big +1! $\endgroup$ – Mark Viola Aug 22 '15 at 23:08

To prove L.G. result, one just needs to apply twice integration by parts, then prove through its favourite technique (for instance, differentiation under the integral sign and computation of a few derivatives of a Beta function) that: $$I_0=\int_{0}^{1}x^2 \log(x)\,\frac{dx}{1-x^2}=1-\frac{\pi^2}{8}, $$ $$I_{-}=\int_{0}^{1}x \log(x)\log(1-x)\,\frac{dx}{1-x^2}=\frac{\pi^2\log(4)-5\zeta(3)}{16}, $$ $$I_{+}=\int_{0}^{1}x \log(x)\log(1+x)\,\frac{dx}{1-x^2}=\frac{-\pi^2\log(4)+9\zeta(3)}{16}. $$ More details to come if wanted. Time to go to bed for me.

  • $\begingroup$ Thank you. +1. If you have time for that, it would be nice to see your complete solution. Good night. $\endgroup$ – user153012 Aug 21 '15 at 23:09
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    $\begingroup$ I worked on this and got as far as the second integration by parts. The two offending terms were $$\int_0^1x \log(x)\frac{\log(1\pm x)}{1\mp x}\,dx$$and I abandoned shortly thereafter due to time constraints ... $\endgroup$ – Mark Viola Aug 22 '15 at 7:04

\begin{align} I&=\int_0^1\operatorname{Li}_3(1-x^2)\ dx\overset{IBP}{=}2\int_0^1\frac{x^2\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx\\ &=2\int_0^1\left(\frac1{1-x^2}-1\right)\operatorname{Li}_2(1-x^2)\ dx\\ &=2\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx-2\int_0^1\operatorname{Li}_2(1-x^2)\ dx\tag{1} \end{align} By the OP, the second integral is $\boxed{\frac{\pi^2}{2}-4}$.

To calculate the first integral, we are going to use the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.

$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=1$ and replacing $x$ with $1-x^2$ we get

$$\frac{\operatorname{Li}_{2}(1-x^2)}{1-x^2}=-\int_0^1\frac{\ln(u)}{1-ux+ux^2}\ du$$

Now we can write

\begin{align} \int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx&=-\int_0^1\ln u\left(\int_0^1\frac{dx}{1-ux+ux^2}\right)\ du\\ &=-\int_0^1\ln u\left(\frac{\arctan\sqrt{\frac{u}{1-u}}}{\sqrt{u-u^2}}\right)\ du, \quad \color{red}{\arctan\sqrt{\frac{u}{1-u}}=\arcsin\sqrt{u}=x}\\ &=-4\int_0^{\pi/2}x\ln(\sin x)\ dx=-4\left(\frac7{16}\zeta(3)-\frac{\pi^2}{8}\ln2\right)=\boxed{\frac{\pi^2}{2}\ln2-\frac74\zeta(3)} \end{align}

Plugging the boxed results of the two integrals in $(1)$, we get

$$I=\pi^2\left(\ln 2-1\right)-\frac72\zeta\left(3\right)+8$$


Note: Since $$\arctan x=-\frac{i}{2}\ln\left(\frac{1+ix}{1-ix}\right)$$

Then \begin{align} \arctan\frac{x}{\sqrt{1-x^2}}&=-\frac{i}{2}\ln\left(\frac{1+\frac{ix}{\sqrt{1-x^2}}}{1-\frac{ix}{\sqrt{1-x^2}}}\right)\\ &=-\frac{i}{2}\ln\left(\frac{\sqrt{1-x^2}+ix}{\sqrt{1-x^2}-ix}*\color{red}{\frac{\sqrt{1-x^2}+ix}{\sqrt{1-x^2}+ix}}\right)\\ &=-\frac{i}{2}\ln\left(\frac{(\sqrt{1-x^2}+ix)^2}{1}\right)\\ &=-i\ln\left(\sqrt{1-x^2}+ix\right)\\ &=\arcsin x \end{align}

and if we replace $x$ with $\sqrt{x}$, we get



Here is a different proof:

Since $$\frac{d}{dy}\arctan\frac{y}{\sqrt{1-y^2}}=\frac1{\sqrt{1-y^2}}$$

Then $$\left.\arctan\frac{y}{\sqrt{1-y^2}}\right|_0^x=\int_0^x\frac1{\sqrt{1-y^2}}\ dy$$

$$\arctan\frac{x}{\sqrt{1-x^2}}=\arcsin x$$


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