First I want to define with the Stirling numbers of the first kind $\left[ \begin{array}{c} n \\ k \end{array} \right]$ a special generalization of the Riemann Zeta function :
$$\zeta_n(m):=\sum\limits_{k=1}^\infty \frac{1}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
and
$$\eta_n(m):=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
which are convergent for the integer values $\enspace m\geq 2$ .
For $\enspace n=0\enspace$ we have $\enspace\zeta_0(m)=\zeta(m)\enspace$ and $\enspace\eta_0(m)=\eta(m)\enspace$ .
Note: Obviously (because of the other results) these series can be expressed by sums of the polylogarithm function and modifications of that.
Please also see here, part Expansion by harmonic numbers, with $\enspace\displaystyle w(n,m):=\frac{m!}{(n-1)!}\left[ \begin{array}{c} n \\ {m+1} \end{array} \right]\enspace$ and it's recursion formula.
Secondly, an extension of an integral as a series, $n\in\mathbb{N}_0$ and $z\in\mathbb{R}\setminus \{2\mathbb{N}\}$ and $nz>-1$:
$ \displaystyle \int\limits_0^\pi x^n \left(2\sin\frac{x}{2}\right)^z dx=i^{-z} \int\limits_0^\pi x^n e^{i\frac{xz}{2}}(1- e^{-ix})^z dx= e^{-i\frac{\pi z}{2}} \int\limits_0^\pi x^n \sum\limits_{k=0}^\infty\binom{z}{k}(-1)^k e^{-ix(\frac{z}{2}-k)} dx$
$\displaystyle =\int\limits_0^\pi x^n e^{i(x-\pi)\frac{z}{2}} dx+ \sum\limits_{v=0}^n \frac{(-1)^v\pi^{n-v} n!}{i^{v+1}(n-v)!} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}} $
$\displaystyle \hspace{3.5cm} -i^{n-1}n!e^{-i\frac{\pi z}{2}} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{ (-1)^k}{(\frac{z}{2}-k)^{n+1}}$
using the main branch of the logarithm and therefore $\displaystyle i=e^{i\frac{\pi}{2}}$ .
The Stirling numbers of the first kind are usually defined by $\enspace \displaystyle \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) $ .
Because of $\enspace \displaystyle (\sum\limits_{v=0}^\infty x^v \frac{d^k}{dz^k}\binom{z}{v}) |_{z=0} =\frac{d^k}{dz^k}(1+x)^z |_{z=0} =(\ln(1+x))^k=k!\sum\limits_{v=k}^\infty (-1)^{v-k} \left[\begin{array}{c} v \\ k \end{array} \right] \frac{x^v}{v!}$
we get $\enspace \displaystyle \binom{z}{k}|_{z=0}=0^k\enspace$ , $\enspace \displaystyle \frac{d}{dz} \binom{z}{k} |_{z=0} = (-1)^{k-1} \left[\begin{array}{c} k \\ 1 \end{array} \right] \frac{1}{k!}= \frac{(-1)^{k-1}}{k} \enspace$ , $\enspace \displaystyle \frac{d^2}{dz^2} \binom{z}{k} |_{z=0} = (-1)^{k-2} \left[\begin{array}{c} k \\ 2 \end{array} \right] \frac{2!}{k!}= \frac{(-1)^k 2}{k}\sum\limits_{j=1}^{k-1}\frac{1}{j} \enspace$ and $\enspace \displaystyle \frac{d^3}{dz^3} \binom{z}{k} |_{z=0} = (-1)^{k-3} \left[\begin{array}{c} k \\ 3 \end{array} \right] \frac{3!}{k!}= \frac{(-1)^{k-1} 3}{k}( (\sum\limits_{j=1}^{k-1}\frac{1}{j})^2 - \sum\limits_{j=1}^{k-1}\frac{1}{j^2} ) $ .
For $(n;k):=(3;3)$ follows
$\displaystyle \int\limits_0^\pi x^3 \left(\ln\left(2\sin\frac{x}{2} \right)\right)^3 dx =$
$\hspace{2cm}\displaystyle =\frac{9\pi^2}{2}\left(\zeta(5)+3\eta(5)-4\eta_1(4)+2\eta_2(3)\right) $
$\hspace{2.5cm}\displaystyle - 90\left(\zeta(7)+\eta(7)\right) +72\left(\zeta_1(6)+\eta_1(6)\right) - 18\left(\zeta_2(5)+\eta_2(5)\right) $
Note:
For the calculations I have used $\enspace\displaystyle\int\limits_0^\pi x^n e^{iax}dx = \frac{(-1)^{n+1} n!}{(ia)^{n+1}}+e^{i\pi a}\sum\limits_{v=0}^n\frac{(-1)^v \pi^{n-v}n!}{(ia)^{v+1}(n-v)!}$
with $\enspace\displaystyle a=-(\frac{z}{2}-k)$ .
And it was necessary to calculate $\enspace\displaystyle\frac{d^m}{dz^m} \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}}|_{z=0}\enspace$ and $\enspace\displaystyle\frac{d^m}{dz^m} e^{-i\frac{\pi z}{2}}\binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{n+1}}|_{z=0}\enspace$ for $\enspace m\in\{0,1,2,3\}$ .
Cross-posting this integral on AoPS brings Y. Sharifi's solution here after a day. Quite amazing one!
I will copy here his entire solution:
Let $I$ be your integral. Using the identity $\ln x \ln(1-x)+\text{Li}_2(x)=\zeta(2)-\text{Li}_2(1-x),$ we have
$$I=-\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x(1-x)} \ dx + \zeta(2)\int_0^1 \left(\frac{\text{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}+\frac{\text{Li}_2(1-x)}{1-x}\right)dx.$$
Let
$$J=\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x(1-x)} \ dx, \ \ \ \ \ K:=\int_0^1 \left(\frac{\text{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}+\frac{\text{Li}_2(1-x)}{1-x}\right)dx.$$
So
$$I=\zeta(2)K - J. \ \ \ \ \ \ \ \ \ (1)$$
We first show that $K=0.$ Start with using integration by parts in $K,$ with $u=\frac{\text{Li}_2(x)}{x}-\zeta(2)+\text{Li}_2(1-x)$ and $dv=\frac{dx}{1-x}.$ Then
$$K=\int_0^1 \ln(1-x)\left(\frac{\ln x}{1-x}-\frac{\ln(1-x)}{x^2}-\frac{\text{Li}_2(x)}{x^2}\right)dx. \ \ \ \ \ \ \ \ \ \ (2)$$
Using the Maclaurin series of $\ln(1-x),$ we quickly find the first integral in $K$
$$\int_0^1 \frac{\ln x \ln(1-x)}{1-x} \ dx = \int_0^1 \frac{\ln x \ln(1-x)}{x} \ dx=\zeta(3). \ \ \ \ \ \ \ \ \ \ (3)$$
Next, we ignore the second integral in $K$ for now and we look at the third one, i.e. $\int_0^1 \frac{\ln(1-x) \text{Li}_2(x)}{x^2} \ dx.$ In this integral, we use integration by parts with $u=\ln(1-x)\text{Li}_2(x)$ and $dv=\frac{dx}{x^2};$ notice that we need to choose $v=1-\frac{1}{x}.$ So
$$\int_0^1 \frac{\ln(1-x) \text{Li}_2(x)}{x^2} \ dx=\int_0^1\left(1-\frac{1}{x}\right)\left(\frac{\text{Li}_2(x)}{1-x}+\frac{\ln^2(1-x)}{x}\right) dx$$
$$=-\int_0^1 \frac{\text{Li}_2(x)}{x} \ dx + \int_0^1 \frac{\ln^2(1-x)}{x} \ dx - \int_0^1 \frac{\ln^2(1-x)}{x^2} \ dx=-\zeta(3)+\int_0^1 \frac{\ln^2x}{1-x} \ dx -\int_0^1 \frac{\ln^2(1-x)}{x^2} \ dx.$$
$$=\zeta(3)-\int_0^1 \frac{\ln^2(1-x)}{x^2} \ dx. \ \ \ \ \ \ \ \ \ (4)$$
Thus, by $(2),(3)$ and $(4),$ we have $K=0$ and hence, by $(1),$
$$I=-J=-\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x(1-x)} \ dx=-2\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x} \ dx.$$
So integration by parts with $u=\text{Li}_2(1-x)$ and $dv=\frac{\text{Li}_2(x)}{x} \ dx$ gives
$$I=2\int_0^1 \frac{\text{Li}_3(x)\ln x}{1-x} \ dx=2\int_0^1 \text{Li}_3(x) \ln x \sum_{m \ge 1}x^{m-1} dx=2\sum_{m \ge 1} \int_0^1 x^{m-1}\text{Li}_3(x) \ln x \ dx$$
$$=2\sum_{m \ge 1} \int_0^1x^{m-1}\sum_{n \ge 1} \frac{x^n}{n^3} \ln x \ dx=2\sum_{m,n \ge 1} \frac{1}{n^3}\int_0^1x^{n+m-1}\ln x \ dx=-2\sum_{m,n \ge 1} \frac{1}{n^3(n+m)^2}$$
$$=-\sum_{m,n \ge 1} \left(\frac{1}{n^3(n+m)^2}+\frac{1}{m^3(n+m)^2}\right). \ \ \ \ \ \ \ \ \ (5)$$
So $(5)$ and the following identity
$$\frac{1}{n^3(n+m)^2}+\frac{1}{m^3(n+m)^2}=\frac{1}{n^3m^2}-\frac{2}{n^2m^3}+\frac{3}{m^3n(n+m)}$$
together give
$$I=-\sum_{m,n \ge 1}\left(\frac{1}{n^3m^2}-\frac{2}{n^2m^3}+\frac{3}{m^3n(n+m)}\right)=\zeta(2)\zeta(3)-3\sum_{m,n \ge 1} \frac{1}{m^3n(n+m)}$$
$$=\zeta(2)\zeta(3)-3\sum_{m \ge 1} \frac{1}{m^4} \sum_{n \ge 1}\left(\frac{1}{n}-\frac{1}{n+m}\right)=\zeta(2)\zeta(3)-3\sum_{m \ge 1} \frac{H_m}{m^4}, \ \ \ \ \ \ \ \ \ (6)$$
where, as usual, $H_m:=\sum_{j=1}^m \frac{1}{j}$ is the $m$-th harmonic number. Now we use Euler's formula
$$\sum_{m \ge 1} \frac{H_m}{m^k}=\left(1+\frac{k}{2}\right)\zeta(k+1)-\frac{1}{2}\sum_{i=1}^{k-2}\zeta(i+1)\zeta(k-i), \ \ \ \ k \ge 2,$$
with $k=4$ to get
$$\sum_{m \ge 1} \frac{H_m}{m^4}=3\zeta(5)-\zeta(2)\zeta(3)$$
and so, by $(6),$
$$I=\zeta(2)\zeta(3)-3(3\zeta(5)-\zeta(2)\zeta(3))=4\zeta(2)\zeta(3)-9\zeta(5).$$
Edit.
This integral was proposed two years ago in RMM and it appeared as problem UP $089$.
See in this link, at the page $70$.
Best Answer
First, we need a preliminary result:
First preliminary result :
$$\ln\left(2 \sinh\left(\frac{x}{2}\right)\right)=\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \tag{1}$$
Proof:
$$ \begin{aligned} \ln\left( \sinh(x)\right)&=\ln\left( \frac{1}{2} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+\ln\left( e^{x}-e^{-x} \right)\\ &=-\ln 2+\ln\left( \frac{e^{-x}}{e^{-x}} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+x+\ln\left( 1-e^{-2x} \right)\\ &=-\ln 2+x-\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n} \qquad \blacksquare \end{aligned} $$
Letting $x \to \frac{x}{2}$ completes the proof
Than:
$$ \begin{aligned} \sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^3}&=-2\int_0^{2\ln(\phi)}x\ln\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx \\ &=-2\int_0^{2\ln(\phi)}x\left(\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \right)\,dx & \left( \text{by eq. (1)}\right)\\ &=-\int_0^{2\ln(\phi)}x^2\,dx+2\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{2\ln(\phi)} xe^{-kx}\,dx\\ &=-\frac{8}{3}\ln^3(\phi)+2\sum_{k=1}^{\infty}\frac{1}{k}\left(-\frac{2\ln(\phi)\phi^{-2k}}{k}+\frac{1}{k}\int_0^{2\ln(\phi)} e^{-kx}\,dx \right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^2}+2\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{k^2}-\frac{(\phi^{-2})^k}{k^2}\right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})+2\zeta(3)-2\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^3}\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})-2\operatorname{Li}_3(\phi^{-2})+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\left( \frac{\pi^{2}}{15}-\ln ^{2} \phi\right)-2\left(\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15} \right)+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)+4\ln^3(\phi)-\frac{4}{3}\ln^3(\phi)-\frac85\zeta(3)+2\zeta(3)\\ &=\frac25\zeta(3) \qquad \blacksquare \end{aligned} $$
Which is the representation of $\zeta(3)$ that Apery used to prove the irrationality of $\zeta(3)$.
Note that we used
$\mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) =\frac{\pi^{2}}{15}-\ln ^{2} \phi$
A proof can be found here in my blog
And:
$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$
Proof of the second relation: Recall the Trilogarithm identity proved here
$ \operatorname{Li}_{3}(x)+\operatorname{Li}_{3}(1-x)+\operatorname{Li}_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}$
And the polylogarithm relation:
$\operatorname{Li}_{n}(x)+\operatorname{Li}_{n}(-x)=2^{1-n}\operatorname{Li}_{n}(x^2)$
Example, letting $x=\phi^{-1}$ in the last relation we obtain
$$\operatorname{Li}_{n}(\phi^{-1})+\operatorname{Li}_{n}(-\phi^{-1})=\frac14\operatorname{Li}_{n}(\phi^{-2})$$
Claim:
$$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$$
Proof:
If we let $x=\phi^{-1}$ in $(2)$ we obtain
$$ \begin{aligned} &\operatorname{Li}_{3}\left(\phi^{-1}\right)+\operatorname{Li}_{3}(1-\phi^{-1})+\operatorname{Li}_{3}\left(1-\frac{1}{\phi^{-1}}\right)=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}-\frac{\ln ^{2}(\phi) \ln (1-\phi^{-1})}{2}\\ &\operatorname{Li}_3\left(\phi^{-1} \right)+\operatorname{Li}_3\left(\phi^{-2} \right)+\operatorname{Li}_3\left(-\phi^{-1} \right)=\zeta(3)-\frac{\ln^3(\phi)}{6}-\frac{\pi^2\ln(\phi)}{6}-\frac{\ln^2(\phi)\ln(\phi^{-2})}{2}\\ &\frac14\operatorname{Li}_{3}\left(\phi^{-2}\right)+\operatorname{Li}_{3}(\phi^{-2})=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}+\ln ^{3}(\phi) \\ &\frac54\operatorname{Li}_{3}\left(\phi^{-2}\right)=\zeta(3)+\frac{5\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6} \\ &\operatorname{Li}_{3}\left(\phi^{-2}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2}\ln (\phi)}{15} \qquad \blacksquare \\ \end{aligned} $$