Justify $\zeta(3)=2\int_0^1 \left(Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x}\right))\log \Gamma(x)dx$ I don't know if this approach to get a formula involving the Apéry constant was in the literature. This idea was a simple idea few minutes ago, when I was studying the answers in this site Math Stack Exchange for the question Integral $\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}$. 
One has that since Wolfram Alpha said that $$\sum_{k=1}^\infty\frac{\cos (2 \pi k x)}{k^2}=\frac{Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x})}{2},$$ 
where $Li_s(z)$ is the polylogarithm function. Then using the dominated convergence theorem we should have 
then $$\zeta(3)=2\int_0^1  \left(Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x}\right))\log \Gamma(x)dx .$$

Question. Please can you justify all these claims to provide us this nice exercise for this site Mathematics Stack Exchange? I say justify the closed-form for the series involving the cosine function (if you find a reference in this site, only is required add it) and after jusfity how one uses the dominated convergence theorem. Thanks in advance.

With respect the use of the theorem, I know that I can bound the cosine function, but how can one bounds  $|\log \Gamma(x) |$ for $0<x<1$?
You can see the calculation from the online calculator with this code 

int_0^1 log (Gamma(x))2(Li_2(e^(-2 i pi x))+Li_2(e^(2 i pi x)))dx.

Notice that the imaginary part is $\approx 0$.
 A: So we have to prove that
$$ \frac{\zeta(3)}{4} = \int_{0}^{1}\sum_{k\geq 1}\frac{\cos(2\pi k x)}{k^2}\log\Gamma(x)\,dx $$
but over the iterval $(0,1)$ the graph of the Fourier cosine series $\sum_{k\geq 1}\frac{\cos(2\pi k x)}{k^2}$ is just the graph of a parabola, since $\sum_{k\geq 1}\frac{\sin(2\pi k x)}{k}$ is a sawtooth wave. Moreover, such a parabola is symmetric with respect to the point $x=\frac{1}{2}$. The claim boils down to:
$$ \frac{\zeta(3)}{4} = \int_{0}^{1/2}\left(\frac{\pi^2}{6}-\pi^2(x-x^2)\right)\log\left(\Gamma(x)\Gamma(1-x)\right)\,dx $$
or, through the reflection formula for the $\Gamma$ function and the substitution $x\to\frac{1}{2}-x$:
$$ \frac{\zeta(3)}{4}=\frac{\pi^2}{12}\int_{0}^{1/2}(12x^2-1)\log\left(\frac{\pi}{\cos(\pi x)}\right)\,dx$$
that is equivalent to:
$$ 3\zeta(3)=\pi^2\int_{0}^{1/2}(1-12x^2)\log(\cos(\pi x))\,dx \tag{1a}$$
or to:
$$ \frac{3\zeta(3)}{\pi^3} = \int_{0}^{1/2}\tan(\pi x)(x-4x^3)\,dx,\tag{1b}$$
$$ \frac{\zeta(3)}{\zeta(2)} = \int_{0}^{1}\frac{\pi}{2}\cot(\pi x/2)x(x-1)(x-2)\,dx.\tag{1c}$$
Since the Weierstrass product of the cosine function gives
$$ \cos(\pi x)=\prod_{n\geq 0}\left(1-\frac{4 x^2}{(2n+1)^2}\right)\tag{2}$$
and
$$ \int_{0}^{1/2}(1-12x^2)\log\left(1-\frac{4 x^2}{(2n+1)^2}\right)\,dx \\= \frac{1}{3}+4 n (1+n)+2 n (1+n) (1+2 n) (\log(n)-\log(n+1)) \tag{3}$$
the whole question boils down to an exercise in summation by parts involving the derivative of the $\zeta$ function (in particular, $\zeta'(-2)$, that is related with $\zeta(3)$ via the reflection formula for the $\zeta$ function). As an alternative, $(1)$ can be checked by expanding both $12x^2-1$ and $\log\cos(\pi x)$ as Fourier series. On the other hand, Chen Wang's solution to the linked question just relies on Kummer's Fourier series for $\log\Gamma$. The Weierstrass product also provides the identities
$$ \frac{\pi}{2}\cot\left(\frac{\pi x}{2}\right)=\frac{1}{x}+\sum_{n\geq 1}\left(\frac{1}{x-2n}+\frac{1}{x+2n}\right) \tag{4}$$
$$ \tan\left(\frac{\pi x}{2}\right)=\frac{4}{\pi}\sum_{n\geq 0}\frac{x}{(2n+1)^2-x^2} \tag{5} $$
that can be used to directly tackle $(1b)$ or $(1c)$.

Now we have a situation that often occurs in mathematics: when proving something, we go through some lemma that is crucial for other situations. In this case we have that the RHS of $(3)$ is summable over $n\geq 1$, so the "smallness" of such term gives us that
$$ \frac{4n(n+1)+\frac{1}{3}}{2n(2n+1)(n+1)} $$
is an excellent approximation of $\log\left(1+\frac{1}{n}\right)$ for any $n\geq 1$, and we may even estimate the error by approximating the associated integral with the Cauchy-Schwarz inequality, integration by parts or other techniques. In our case we get that the approximation error is less than $1.3\cdot 10^{-3}$ for any $n\geq 1$, and 
$$ \log(2)\approx\left(\frac{5}{6}\right)^2. $$
