TITLE: How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
QUESTION [16 upvotes]: @User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi  \Im(\text{Li}_3(1+i))+64 \text{Li}_4\left(\frac{1}{2}\right)-233 \zeta(4)-40  \ln ^2(2)\zeta(2)+\frac{8}{3}\ln ^4(2)}\tag2$$
I was able to prove $(1)$ but had some difficulty proving $(2)$. Any idea?
I am going to show my proof of $(1)$ hoping it helps you prove $(2)$:
We showed in this question that
$$\sum_{n=1}^\infty\frac{4^ny^n}{n^2{2n\choose n}}=2\int_0^y \frac{\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}dx$$
multiply both sides by $\frac{1}{y\sqrt{1-y}}$ then $\int_0^1$ with respect to $y$ and use $\int_0^1\frac{y^{n-1}}{\sqrt{1-y}}dy=\frac{4^n}{n{2n\choose n}}$ we obtain
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=2\int_0^1\int_0^y \frac{\arcsin \sqrt{x}}{y\sqrt{x}\sqrt{1-x}\sqrt{1-y}}dxdy$$
$$=2\int_0^1\frac{\arcsin\sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(\int_x^1\frac{dy}{y\sqrt{1-y}}\right)dx$$
$$=2\int_0^1\frac{\arcsin\sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(2\ln(1+\sqrt{1-x})-\ln x\right)dx$$
$$\overset{\sqrt{x}=\sin \theta}{=}8\int_0^{\pi/2}x\ln(1+\cos x)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$
$$=8\int_0^{\pi/2}x\ln(2\cos^2\frac x2)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$
$$=32\int_0^{\pi/4}x\ln(2\cos^2x)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$
$$=32\underbrace{\int_0^{\pi/4}x\ln(2)dx}_{\frac3{16}\ln(2)\zeta(2)}+64\underbrace{\int_0^{\pi/4}x\ln(\cos x)dx}_{\frac{\pi}{8}\text{G}-\frac3{16}\ln(2)\zeta(2)-\frac{21}{128}\zeta(3)}-8\underbrace{\int_0^{\pi/2}x\ln(\sin x)dx}_{\frac7{16}\zeta(3)-\frac34\ln(2)\zeta(2)}$$
$$=8\pi\text{G}-14 \zeta (3)$$
The last two integrals follow from using the Fourier series of $\ln(\cos x)$ and $\ln(\sin x)$.
All approaches are appreciated. Thank you.

Addendum: Here is an easier way to prove $(1)$:
We have
$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
or
$$\sum_{n=1}^\infty\frac{4^nx^n}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{x})$$
Divide both sides by $x\sqrt{1-x}$ then $\int_0^1$ and use $\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}dx=\frac{4^n}{n{2n\choose n}}$ we have
$$\sum_{n=1}^\infty\frac{16^n}{n^3{2n\choose n}^2}=2\int_0^1\frac{\arcsin^2(\sqrt{x})}{x\sqrt{1-x}}dx$$
$$\overset{\sqrt{x}=\sin x}{=}4\int_0^{\pi/2}x^2 \csc(x)dx$$
$$\overset{IBP}{=}-8\int_0^{\pi/4} x\ln(\tan\frac x2)dx=8\pi\text{G}-14\zeta(3)$$
where the last result follows from the Fourier series of $\ln(\tan\frac x2)$.

REPLY [4 votes]: Since
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
we can write
$$\sum_{n=1}^\infty\frac{4^nx^{n}}{n{2n\choose n}}=\frac{2\sqrt{x}\arcsin \sqrt{x}}{\sqrt{1-x}}$$
Multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^y$ and use the fact that $\int_0^y - x^{n-1}\ln xdx=\frac{1}{n^2}y^n-\frac{\ln y}{n}y^n$
$$\sum_{n=1}^\infty\frac{4^ny^n}{n^3{2n\choose n}}-\ln y\sum_{n=1}^\infty\frac{4^ny^n}{n^2{2n\choose n}}=-\int_0^y \frac{2\ln x\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}dx$$
Next multiply both sides by $\frac{1}{y\sqrt{1-y}}$ then $\int_0^1$ we get
$$\sum_{n=1}^\infty\frac{4^n}{n^3{2n\choose n}}\left(\int_0^1\frac{y^{n-1}}{\sqrt{1-y}}dy\right)-\int_0^y\frac{\ln y}{y\sqrt{1-y}}\left(\sum_{n=1}^\infty\frac{(2\sqrt{y})^{2n}}{n^2{2n\choose n}}\right)dx$$
$$=-\int_0^1\int_0^y \frac{2\ln x\arcsin \sqrt{x}}{y\sqrt{x}\sqrt{1-x}\sqrt{1-y}}dxdy=-\int_0^1 \frac{2\ln x\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(\int_x^1\frac{dy}{y\sqrt{1-y}}\right)dx$$
$$=-\int_0^1 \frac{2\ln x\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(2\ln(1+\sqrt{1-x})-\ln x\right)dx$$
$$\overset{\sqrt{x}=\sin\theta}{=}16\int_0^{\pi/2}x\ln(\sin x)\ln\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=16\int_0^{\pi/2}x\ln(\sin x)\ln\left(\tan(\frac x2)\right)dx$$
$$\overset{x\to 2x}{=}64\int_0^{\pi/4}x\ln(\sin(2x))\ln\left(\tan x\right)dx$$
$$=64\int_0^{\pi/4}x[\ln(2)+\ln(\sin x)+\ln(\cos x)][\ln(\sin x)-\ln(\cos x)]dx$$
$$=64\ln(2)\int_0^{\pi/4}x\ln(\tan x)dx+64\int_0^{\pi/4}x\ln^2(\sin x)dx-64\int_0^{\pi/4}x\ln^2(\cos x)dx$$
For the LHS, use $\int_0^1\frac{y^{n-1}}{\sqrt{1-y}}dy=\frac{4^n}{n{2n\choose n}}$ and $\sum_{n=1}^\infty\frac{(2\sqrt{y})^{2n}}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{y})$ we get
$$\text{LHS}=\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}-2\int_0^1\frac{\ln y\arcsin^2(\sqrt{y})}{y\sqrt{1-y}}dy$$
$$\overset{\sqrt{y}=\sin \theta}{=}\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}-8\int_0^{\pi/2} x^2\csc x\ln(\sin x)dx$$
Therefore
$$\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}=64\ln(2)\int_0^{\pi/4}x\ln(\tan x)dx-64\int_0^{\pi/4}x\ln^2(\cos x)dx$$
$$+64\int_0^{\pi/4}x\ln^2(\sin x)dx+8\int_0^{\pi/2} x^2\csc x\ln(\sin x)dx\tag1$$

The first integral can be done via Fourier series:
$$\int_0^{\pi/4} x\ln(\tan x)dx=\frac{7}{16}\zeta(3)-\frac{\pi}{4}\text{G}\tag2$$
The second integral:
$$\int_0^{\pi/4}x\ln^2(\cos x)dx=\int_0^{\pi/2}x\ln^2(\cos x)dx-\underbrace{\int_{\pi/4}^{\pi/2}x\ln^2(\cos x)dx}_{x\to \pi/2-x}$$
$$=\int_0^{\pi/2}x\ln^2(\cos x)dx-\int_{\pi/4}^{\pi/2}(\frac{\pi}{2}-x)\ln^2(\sin x)dx$$
$$=\int_0^{\pi/2}x\ln^2(\cos x)dx-\frac{\pi}{2}\int_0^{\pi/4}\ln^2(\sin x)dx+\int_0^{\pi/4}x\ln^2(\sin x)dx$$
Plugging this result along with $(2)$ in $(1)$, the integral $\int_0^{\pi/4}x\ln^2(\sin x)dx$ nicely cancels out getting:
$$\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}=28\ln(2)\zeta(3)-16\pi\ln(2)\text{G}-64\int_0^{\pi/2}x\ln^2(\cos x)dx$$
$$+32\pi\int_0^{\pi/4}\ln^2(\sin x)dx+8\int_0^{\pi/2} x^2\csc x\ln(\sin x)dx$$
Lets manipulate the first integral using the same trick $x\to \pi/2-x$:
$$\int_0^{\pi/2}x\ln^2(\cos x)dx=\int_0^{\pi/2}(\frac{\pi}{2}-x)\ln^2(\sin x)dx$$
$$=\frac{\pi}{2}\int_0^{\pi/2}\ln^2(\cos x)dx-\int_0^{\pi/2}x\ln^2(\sin x)dx$$
By Beta function we have
$$\frac{\pi}{2}\int_0^{\pi/2}\ln^2(\cos x)dx=\frac{15}{8}\zeta(4)+\frac32\ln^2(2)\zeta(2)$$
and our sum boils down to
$$\sum_{n=1}^\infty\frac{16^n}{n^4{2n\choose n}^2}=28\ln(2)\zeta(3)-16\pi\ln(2)\text{G}-120\zeta(4)-96\ln^2(2)\zeta(2)$$
$$+64\underbrace{\int_0^{\pi/2}x\ln^2(\sin x)dx}_{\mathcal{\Large{I_1}}}+32\pi\underbrace{\int_0^{\pi/4}\ln^2(\sin x)dx}_{\mathcal{\Large{I_2}}}+8\underbrace{\int_0^{\pi/2}x\csc x\ln(\sin x)dx}_{\mathcal{\Large{I_3}}}$$
$\mathcal{I}_1$ is calculated here:
$$\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$
$\mathcal{I}_2$ is calculated here
$$\int_{0}^{\pi /4} \ln^{2}(\sin x) \ dx = \frac{\pi^{3}}{192} + G\frac{  \ln(2)}{2} + \frac{3 \pi}{16} \ln^{2}(2) + \text{Im} \  \text{Li}_{3}(1+i).$$
$\mathcal{I}_3$ is calculated here
$$\int_0^{\pi/2} \frac{x^2 \ln(\sin x)}{\sin (x)} dx=-4 \pi  \Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}-\frac{7}{2} \zeta (3) \ln (2)+\frac{135}{16}\zeta(4)+\frac{3}{4} \zeta(2) \ln ^2(2)$$
$$=4\pi\Im\{\text{Li}_3(1+i)\}-\frac{45}{4}\zeta(4)-\frac72\ln(2)\zeta(3)-\frac32\ln^2(2)\zeta(2)$$
The last result follows from using
$$\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}=\frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2(2)-\Im\{\text{Li}_3(1+i)\}$$
Collecting the three integrals we finally get
$$\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi  \Im\{\text{Li}_3(1+i)\}+64 \text{Li}_4\left(\frac{1}{2}\right)-233 \zeta(4)-40  \ln ^2(2)\zeta(2)+\frac{8}{3}\ln ^4(2)$$

Thanks to Cornel for the hint $x\to \pi/2-x$ which simplifies $\int_0^{\pi/2}x\ln^2(\cos x)dx$ to known integrals.