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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.

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$$I=\int_0^\frac{\pi}{2} x\cot x \ln(\cos x)dx\overset{IBP}=\int_0^\frac{\pi}{2}x\tan x\ln(\sin x)dx-\int_0^\frac{\pi}{2}\ln(\sin x)\ln(\cos x)dx$$ 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. $$J=-\frac12 \int_0^1 \frac{\ln(1+x^2)}{x(1+x^2)}dx-\frac12 \int_0^1 \frac{x\ln(1+x^2)-x\ln (x^2)}{1+x^2}dx$$ $$=-\frac12 \int_0^1 \frac{\ln(1+x^2)}{x}+\int_0^1 \frac{x\ln x}{1+x^2}dx=-\int_0^1 \frac{\ln(1+x^2)}{x}dx=-\frac{\pi^2}{24}$$ $$\Rightarrow I= 4J-\frac\pi4 \ln^22+\frac{\pi^3}{96}=-\frac{\pi}{4}\ln^2 2\Rightarrow P=\frac{\pi^3}{12}-\pi \ln^2 2$$

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    $\begingroup$ Magnificent approach! I had never considered to use IBP like that before. Thanks :) $\endgroup$ – clathratus May 14 at 22:07

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