Proof of $\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}=\frac{17\pi^4}{3240}$

Here is my solution:

Step 1. Reduction to an integral representation

Let $S$ denote the summation in question. Then by the successive application of integration by parts, we obtain

\begin{align*} S &= 8 \int_{0}^{1} \frac{1}{x} \int_{0}^{x/2} \frac{\arcsin^{2} t}{t} \, dt dx = -8\int_{0}^{1} \frac{\arcsin^{2} (x/2)}{x} \log x \, dx \\ &= 8 \int_{0}^{1} \frac{\arcsin (x/2)}{\sqrt{1 - (x/2)^{2}}} \log^{2} x \, \frac{dx}{2}. \end{align*}

Thus with the substitution $x = 2 \sin\theta$, we have

$$ S = 8 \int_{0}^{\frac{\pi}{6}} \theta \log^{2} (2 \sin\theta) \, d\theta. $$

To evaluate this integral, note that for $0 < \theta < \frac{\pi}{6}$ we have

$$ e^{i\theta} \cdot 2 \sin \theta = i \cdot (1 - e^{2i\theta}). $$

Taking logarithm (with the branch cut $(-\infty, 0]$ as usual) to both sides, it follows that

$$ i\theta + \log (2\sin\theta) = \frac{i\pi}{2} + \log(1 - e^{2i\theta}). $$

Cubing both sides and integrating on $\left( 0, \frac{\pi}{6} \right)$ and taking imaginary parts only,

$$ S = \frac{2}{3} \left( \frac{\pi}{6} \right)^{4} + \frac{8}{3} \Im \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta. \tag{1} $$

Step 2. Some complex-analysis techniques

Now we focus on the integral in the imaginary part:

$$ I := \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta \tag{2}. $$

Once we evaluate the imaginary part of $I$, the identity $(1)$ immediately gives us the answer. We first make the substitution $z = 1 - e^{2i\theta}$ and $\omega = e^{-i\pi/3}$ to obtain

$$ I = \int_{0}^{\omega} \left( \frac{i\pi}{2} + \log z \right)^{3} \frac{dz}{2i(z-1)}. $$

Here, the path of integration is a circular arc joining from $0$ to $\omega$ centered at $1$ (green-colored path in the figure below).

But since the integrand is analytic for $0 < \Re z < 1$, we may change the path of integration as $z = \omega t$ for $0 \leq t \leq 1$ (blue-colored path in the figure above). This gives

$$ I = \frac{1}{2i} \int_{0}^{1} \left( \frac{i\pi}{6} + \log t \right)^{3} \frac{\omega \, dt}{\omega t - 1}. $$

Plugging $t = e^{-x}$, $I$ reduces to

\begin{align*} I &= \frac{1}{2} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} \frac{\omega e^{-x}}{1 - \omega e^{-x}} \, dx = \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} e^{-nx} \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \left( -\frac{6 i}{n^4}-\frac{\pi }{n^3}+\frac{i \pi ^2}{12 n^2}+\frac{\pi ^3}{216 n} \right). \end{align*}

Taking the imaginary part,

\begin{align*} \Im I &= -3 \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^4} + \frac{\pi}{2} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n^3} + \frac{\pi^2}{24} \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^2} - \frac{\pi^3}{432} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n}. \tag{3} \end{align*}

Step 3. Evaluation of a series

Note that for $0 < \theta < \pi$, we have

$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \Im \sum_{n=1}^{\infty} \frac{e^{in\theta}}{n} = - \Im \log(1 - e^{i\theta}) = \frac{\pi-\theta}{2}. $$

Integrating both sides, we obtain

$$ \sum_{n=1}^{\infty} \frac{1-\cos n\theta}{n^{2}} = \frac{\theta (2 \pi -\theta )}{4} \quad \Longrightarrow \quad \sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{2}} = \frac{\theta ^2}{4}-\frac{\pi \theta }{2}+\frac{\pi ^2}{6}.$$

Repeating this procedure, we obtain

$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^{3}} = \frac{\theta ^3}{12}-\frac{\pi \theta ^2}{4}+\frac{\pi ^2 \theta }{6}$$

and

$$ \sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{4}} = -\frac{\theta ^4}{48}+\frac{\pi \theta ^3}{12}-\frac{\pi ^2 \theta ^2}{12}+\frac{\pi^4}{90}.$$

Plugging $\theta = \frac{\pi}{3}$, we have

\begin{align*} \sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{2}} &= \frac{\pi}{3} \\ \sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{2}} &= \frac{\pi^3}{36} \\ \sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{3}} &= \frac{5 \pi^3}{162} \\ \sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{4}} &= \frac{91 \pi ^4}{19440} \end{align*}

Plugging these to $(3)$, we have

$$ \Im I = \frac{23 \pi^4}{12960} \quad \Longrightarrow \quad S = \frac{17 \pi^4}{3240} = \frac{17}{36}\zeta(4) $$

as desired.

Your question is the topic of a paper I recently submitted here: https://www.researchgate.net/publication/338188132_On_Central_Binomial_Series_Related_to_z4 (it will be available on arXiv in a couple of days, and I will update the link as soon as it is available). It is too long to write down as a complete answer since I invoke a lot of preliminary lemmas. In particular, I give an alternative proof of the this central binomial series identity using the double integral \begin{align*} K &= \int_0^1 \int_{0}^{1-x} \frac{2\log^2(x+y)}{1-xy} \ dy \ dx. \end{align*} On one hand $K,$ turns out to be equal to @Sangchul Lee's integral representation of the series upon a change of variables $x=(u+v)/2, y=(u-v)/2.$ But integrating the original representation of $K$ directly gives a sum of $3$ polylogarithmic integrals, which can be evaluated with some work.