Evaluate $\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}$ In this  post give the result of $$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)$$
Then i wonder if i change $n^3$ to $(2n+1)^3$ so what is the result. I check with CAS give the result: $$\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}=-\pi\text{G}+\frac{7}{2} \zeta (3)\tag1$$.
My attemp is: $\frac{2}{(2n+1)^3}=\int_{0}^{1}x^{2n}\log^{2}{x}dx$, then (1) become: $\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}=\frac{1}{2}\sum _{n=0}^{\infty } \frac{16^n}{\binom{2 n}{n}^2}\int_{0}^{1}x^{2n}\log^{2}{x}dx=\int_{0}^{1}\log^{2}{x}dx\sum _{n=0}^{\infty } x^{2n}\frac{16^n}{\binom{2 n}{n}^2}$. But the sum is seem harder. I don't have much experience with central binomial coefficient. Can you give some hints? Thank you.
 A: Following Jack D' Aurizio's suggestion from the comments, we start off with the well-known infinite series expansion for $\arcsin x$
$$\frac {\arcsin x}{\sqrt{1-x^2}}=\sum\limits_{n=0}^{+\infty}\frac {4^nx^{2n+1}}{(2n+1)\binom {2n}n}$$
The right-hand side looks somewhat close to the infinite series under question, with the main difference being the $4^n$ in the numerator and the linear $2n+1$ term and binomial in the denominator. To transform the right-hand side into the desired form, divide both sides by $x$ and integrate with respect to $x$. This will add an extra $2n+1$ term in the denominator
$$\sum\limits_{n=0}^{+\infty}\frac {4^nx^{2n+1}}{(2n+1)^2\binom {2n}n}=\int\frac {\arcsin x}{x\sqrt{1-x^2}}\,\mathrm dx\tag{1}$$
Next, make the substitution $x=\sin\theta$. The purpose of this substitution is to exploit another well-known integral identity of $\sin\theta$
$$\int\limits_0^{\pi/2}\sin^{2n+1}\theta\,\mathrm d\theta=\frac {\sqrt\pi\Gamma(n+1)}{(2n+1)\Gamma\left(n+\frac 12\right)}=\frac {4^n}{(2n+1)\binom {2n}n}$$
Upon integrating $(1)$, the left-hand side becomes the infinite series
$$\sum\limits_{n=0}^{+\infty}\frac {16^n}{(2n+1)^3\binom {2n}n^2}=\Re\int\limits_0^{\pi/2}\theta\log\left(\frac {1-e^{i\theta}}{1+e^{i\theta}}\right)\,\mathrm d\theta+\Im\int\limits_0^{\pi/2}\operatorname{Li}_2\left(-e^{i\theta}\right)-\operatorname{Li}_2\left(e^{i\theta}\right)\,\mathrm d\theta\tag{2}$$
The first integral can be evaluated by splitting apart the natural logarithm term and then using integration by parts on $u=x$ and $v=i\operatorname{Li}_2\left(\mp e^{ix}\right)$
$$\int\limits_0^{\pi/2}x\log\left(1\pm e^{ix}\right)\,\mathrm dx=\frac {\pi i}2\operatorname{Li}_2(\mp i)-\operatorname{Li}_3(\mp i)+\zeta(3)$$
Subtracting the two integrals gives
$$\int\limits_0^{\pi/2}x\log\left(\frac {1-e^{ix}}{1+e^{ix}}\right)\,\mathrm dx=\frac {\pi i}2\operatorname{Li}_2(i)-\frac {\pi i}2\operatorname{Li}_2(-i)+\operatorname{Li}_3(-i)-\operatorname{Li}_3(i)\tag{3}$$
The complex polylogarithmic functions can be further simplified by using their respective infinite-sum definition and collecting the real and imaginary components. For instance, with $\operatorname{Li}_2(i)$, we have
$$\operatorname{Li}_2(i)=\sum\limits_{n=1}^{+\infty}\frac {(-1)^n}{(2n)^2}+i\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}{(2n-1)^2}=-\frac {\pi^2}{48}+iG$$
Where $G$ is Catalan's constant. Repeating the same process with the other logarithms, we have the following
$$\begin{align*}\operatorname{Li}_2(\pm i) & =-\frac {\pi^2}{48}\pm iG\tag{4}\\\operatorname{Li}_3(\pm i) & =-\frac 3{32}\zeta(3)\pm\frac {\pi^3 i}{32}\tag{5}\end{align*}$$
Hence
$$\begin{align*}\Re\int\limits_0^{\pi/2}x\log\left(1-e^{ix}\right)\,\mathrm dx & =\frac {35}{32}\zeta(3)-\frac {\pi G}2\\\Re\int\limits_0^{\pi/2}x\log\left(1+e^{ix}\right)\,\mathrm dx\, & =\frac {\pi G}2-\frac {21}{32}\zeta(3)\end{align*}$$
Taking the real part of equation $(3)$, then
$$\Re\int\limits_0^{\pi/2}x\log\left(\frac {1-e^{ix}}{1+e^{ix}}\right)\,\mathrm dx=-\pi G+\frac 74\zeta(3)\tag{6}$$
The latter two integrals can both be solved using substitution
$$\begin{align*}\int\limits_0^{\pi/2}\operatorname{Li}_2\left(-e^{ix}\right)\,\mathrm dx & =-\frac 34\zeta(3)i-i\operatorname{Li}_3(-i)\tag{7}\\\int\limits_0^{\pi/2}\operatorname{Li}_2\left(e^{ix}\right)\,\mathrm dx & =\zeta(3) i-i\operatorname{Li}_3(i)\tag{8}\end{align*}$$
Putting everything together, then we get
$$\begin{align*}\sum\limits_{n=0}^{+\infty}\frac {16^n}{(2n+1)^3\binom {2n}n^2} & =\frac 74\zeta(3)-\pi G+\frac 34\zeta(3)+\zeta(3)\\ & =\frac 72\zeta(3)-\pi G\end{align*}$$
A: Other than @Frank W's elegant solution, here is a different approach.
It is well-known that:
$\displaystyle \tag*{}\frac {\arcsin x}{\sqrt{1-x^2}}=\sum\limits_{n\ge 0}\frac {4^nx^{2n+1}}{(2n+1)\binom {2n}n}$
Dividing by $x$ and doing $\int_0^x \,dx$ on both sides gives:
$\displaystyle \tag{1}\int_0^x\frac {\arcsin x}{x\sqrt{1-x^2}}dx=\sum\limits_{n\ge 0}\frac {4^nx^{2n+1}}{(2n+1)^2\binom {2n}n}$
Now, we use the fact that
$$  \int_0^{\pi/2} \sin^{2n+1}(\theta) \, d\theta = \frac {4^n}{(2n+1)\binom {2n}n} $$
This tells us that if we let $x = \sin (\theta)$ and taking $\int_0^{\pi /2} \, d\theta$ both sides of $(1)$ gives our sum. Thus,
$$ \sum\limits_{n\ge 0}\frac {16^n}{(2n+1)^3\binom {2n}n} = \underbrace{\int_0^{\pi /2} \int_0^\theta \frac{x}{\sin x} \, dx \, d\theta}_{I}$$
Integrating by parts with $u=x$ and $v' = \log(\tan\frac x2)$, we get
$$I = \int_0^{\pi /2} \theta \log \tan \frac \theta 2 \, d\theta -\int_0^{\pi/2}\int_0^\theta \log \tan \frac x2 \, dx \, d\theta $$
Now, with the help of Fourier series, we have $\forall \ x \in (0, \pi)$
$$\begin{align} \tag{2} \log \sin \frac x2 &= -\log 2 - \sum_{n \ge 1} \frac{\cos nx}{n} \\ \tag{3}\log \cos \frac x2 &= -\log 2 + \sum_{n \ge 1} \frac{(-1)^{n-1}\cos nx}{n}\end{align}$$
Doing $(2)-(3)$ gives the series for $\log \tan \frac x2$. Now,plugging in the value, we get
$$ \begin{align}  I &= \frac \pi 2\sum_{n\ge 1} \frac{(-1)^n \sin (n \pi/2)}{n^2} - \underbrace{\sum_{n\ge 1} \frac{(-1)^n}{n^3}}_{\dfrac{-3\zeta(3)}{4}} - \frac \pi 2 \sum_{n \ge 1} \frac{\sin (n \pi /2)}{n^2} + \underbrace{\sum_{n \ge 1} \frac{1}{n^3}}_{\displaystyle \zeta(3)} + \underbrace{2\sum_{n \ge 1} \frac{1}{(2n+1)^3} + 2}_{\dfrac{7 \zeta(3)}{4}} \\\\ &= \frac{7 \zeta(3)}{2} - \frac \pi 2 \sum_{n \ge 1} \frac{\sin(n\pi /2)((-1)^n-1)}{n^2} \\ &=  \frac{7 \zeta(3)}{2} -  \pi \sum_{n \ge 1} \frac{(-1)^n}{(2n+1)^2} \\ &= {\color{Red}{\frac{7 \zeta(3)}{2} - \pi G}}\end{align}$$
