Integral: $\int_0^1\frac{\mathrm{Li}_2(x^2)}{\sqrt{1-x^2}}dx$

Question

I am trying to evaluate $$P=\frac\pi2\sum_{n\geq1}\frac{{2n\choose n}}{4^n n^2}$$ I used the beta function to show that $$P=\int_0^1\frac{\mathrm{Li}_2(x^2)}{\sqrt{1-x^2}}dx$$ IBP: $$P=\sin^{-1}(x)\mathrm{Li}_2(x^2)\big|_0^1+2\int_0^1\frac{\ln(1-x^2)}{x}\sin^{-1}(x)dx$$ Which is $$P=\frac{\pi^3}{12}+4\int_0^{\pi/2}x\cot(x)\ln(\cos x)dx$$ Which I'm not sure how to handle. I will continue working on this integral and update on my progress.

Answer 1

$$I=\int_0^\frac{\pi}{2} x\cot x \ln(\cos x)dx\overset{IBP}=\int_0^\frac{\pi}{2}x\tan x\ln(\sin x)dx-\int_0^\frac{\pi}{2}\ln(\sin x)\ln(\cos x)dx$$ The substitution $\frac{\pi}{2}-x=x$ in the first integral gives: $$ I=\frac{\pi}{2} \int_0^\frac{\pi}{2}\cot x\ln(\cos x)dx-I-\int_0^\frac{\pi}{2}\ln(\sin x)\ln(\cos x)dx$$ $$I=\frac{\pi}{4} \int_0^\frac{\pi}{2}\cot x\ln(\cos x)dx -\frac\pi4 \ln^22+\frac{\pi^3}{96}$$ I won't focus on the second integral since I believe there is a way to avoid all the calculation and magically simplify it, but here is an approach. $$J=\int_0^\frac{\pi}{2}\cot x\ln(\cos x)dx\overset{\tan x=t}=-\frac12 \int_0^\infty \frac{\ln(1+x^2)}{x(1+x^2)}dx$$ Split the integral in the point $1$ then let $\frac{1}{x}\to x$ in the second part. $$J=-\frac12 \int_0^1 \frac{\ln(1+x^2)}{x(1+x^2)}dx-\frac12 \int_0^1 \frac{x\ln(1+x^2)-x\ln (x^2)}{1+x^2}dx$$ $$=-\frac12 \int_0^1 \frac{\ln(1+x^2)}{x}+\int_0^1 \frac{x\ln x}{1+x^2}dx=-\int_0^1 \frac{\ln(1+x^2)}{x}dx=-\frac{\pi^2}{24}$$ $$\Rightarrow I= 4J-\frac\pi4 \ln^22+\frac{\pi^3}{96}=-\frac{\pi}{4}\ln^2 2\Rightarrow P=\frac{\pi^3}{12}-\pi \ln^2 2$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} P & \equiv \bbox[5px,#ffd]{% {\pi \over 2}\sum_{n\ \geq\ 1}{{2n \choose n} \over 4^{n}n^{2}}} = {\pi \over 2}\sum_{n = 1}^{\infty} {{-1/2 \choose n}\pars{-4}^{n} \over 4^{n}} \bracks{-\int_{0}^{1}\ln\pars{x}x^{n - 1}\,\dd x} \\[5mm] & = -\,{\pi \over 2}\int_{0}^{1}\ln\pars{x} \bracks{\sum_{n = 1}^{\infty}{-1/2 \choose n}\pars{-x}^{n}}\,{\dd x \over x} \\[5mm] & = -\,{\pi \over 2}\int_{0}^{1}\ln\pars{x} \bracks{\pars{1 - x}^{-1/2} - 1}\,{\dd x \over x} \\[5mm] = &\ \left. -\,{\pi \over 2}\,\partiald{}{\mu}\int_{0}^{1}x^{\mu - 1} \bracks{\pars{1 - x}^{-1/2} - 1}\,\dd x \,\right\vert_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ -\,{\pi \over 2}\,\partiald{}{\mu}\bracks{% {\Gamma\pars{\mu}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 1/2}} - {1 \over \mu}}_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ -\,{\pi \over 2}\,\partiald{}{\mu}\braces{{1 \over \mu}\bracks{% {\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 1/2}} - 1}}_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ -\,{\pi \over 4}\,\partiald[2]{}{\mu}\bracks{% {\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 1/2}}}_{\ \mu\ =\ 0^{+}} = \bbx{{\pi^{3} \over 12} - \pi\ln^{2}\pars{2}} \\ & \end{align}

Answer 3

Using 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 $x^2$ we get

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

we can write \begin{align} I&=\int_0^1\frac{\operatorname{Li}_{2}(x^2)}{\sqrt{1-x^2}}\ dx=-\int_0^1\ln u\left(\int_0^1\frac{x^2}{(1-ux^2)\sqrt{1-x^2}}\ dx\right)\ du\\ &=-\frac{\pi}{2}\int_0^1\frac{\ln u}{u}\left(\frac{1}{\sqrt{1-u}}-1\right)\ du\overset{IBP}{=}\frac{\pi}{8}\int_0^1\ln^2u(1-u)^{-3/2}\ du\\ &=\frac{\pi}{8}\frac{\partial^2}{\partial\alpha^2}\lim_{\alpha\ \mapsto1}\text{B}\left(\alpha,-\frac12\right)=\frac{\pi}{8}\left(\frac23\pi^2-8\ln^22\right)=\frac{\pi^3}{12}-\pi\ln^22 \end{align}

Answer 4

By courtesy of @JanG's answer here, there's yet another method which I mentioned in the comments above.

As $\ln(ix+1)=\frac{\ln(x^2+1)}{2}+i\arctan x$,

\begin{align*} I&=\int_0^{\pi/2}\underbrace{x\cot x\ln\cos x\,dx}_{\tan x\mapsto x}\\ &=\int_0^\infty \left(\arctan x\right)\left(\frac{1}{x}\right)\left(\frac{\ln(x^2+1)}{-2}\right)\left(\frac{dx}{x^2+1}\right)\\ &=-\frac{1}{4}\int_{-\infty}^\infty \frac{\arctan x\ln(x^2+1)}{x(x^2+1)}dx\\ &=-\frac{1}{4}\int_{-\infty}^\infty \frac{\mathfrak{Im}\left(\ln^2(ix+1)\right)}{x(x^2+1)}dx \end{align*}

Then by contour integration along a semicircle contour on the lower-half plane,

\begin{align*} \require{cancel} I&=-\frac{1}{4}\mathfrak{Im}\left(-2\pi i\mathop{\rm Res}_{x=-i}\frac{\ln^2(ix+1)}{x(x^2+1)}+\cancelto{0}{(\text{arc})}\right)\\ &=-\frac{1}{4}\mathfrak{Im}\left(-2\pi i\frac{\ln^2 2}{(-i)(-2i)}\right)\\ &=-\frac{\pi}{4}\ln^2 2 \end{align*}

And hence the desired value of $P$.