# Closed-form of $\int_0^1 \operatorname{Li}_3\left(1-x^2\right) dx$

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.

• 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. Aug 21, 2015 at 21:44
• 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 Aug 21, 2015 at 23:12

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}

• Well Done! A big +1! Aug 22, 2015 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.

• Thank you. +1. If you have time for that, it would be nice to see your complete solution. Good night. Aug 21, 2015 at 23:09
• 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 ... Aug 22, 2015 at 7:04

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

• Nice but how did you arrived here sir ? Aug 21, 2015 at 22:43
• @RenatoFaraone Namagiri was actually looking for Kleo, but found only L.G. Aug 21, 2015 at 22:48
• 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 $\frac{x\log x}{1-x^2}$. Aug 21, 2015 at 22:56
• Using differentiation under the integral sign, that boils down to evaluating a few derivatives of a Beta function, not that hard. Aug 21, 2015 at 22:59
• @JackD'Aurizio Feel free to post your solution. Thank you L.G. for your answer. +1. Aug 21, 2015 at 23:00

\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

$$\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx=-\int_0^1\ln u\left(\int_0^1\frac{dx}{1-ux+ux^2}\right)\ du$$

$$=-\int_0^1\ln u\left(\frac{\arctan\sqrt{\frac{u}{1-u}}}{\sqrt{u-u^2}}\right)\ du, \quad \color{red}{\arctan\sqrt{\frac{u}{1-u}}=\arcsin\sqrt{u}=x}$$

$$=-4\int_0^{\pi/2}x\ln(\sin x)\ dx=-4\left(\frac7{16}\zeta(3)-\frac{\pi^2}{8}\ln2\right)=\boxed{\frac{\pi^2}{2}\ln2-\frac74\zeta(3)}$$

where the last result follows from the Fourier series of $$\ln(\sin x)=-\ln2-\sum_{n=1}^\infty \frac{(-1)^n \cos(2nx)}{n}$$.

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

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

Here is a different proof:

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

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

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