Prove $\int_0^\infty\Big|\Gamma\Big(\frac13+\frac{i\alpha}{\pi}\Big)\Big|^6\alpha \sinh\alpha~d\alpha=\frac{\pi\sqrt{27}}{4}\Gamma^9\Big(\frac23\Big)$ In the book Formfactors in integrable models of quantum field theory F.A. Smirnov gives the following integral (there is a typo in the book, kindly corrected by Mr. Smirnov)
\begin{align}
&\int_{-\infty}^{\infty}\prod_{j=1}^3\Gamma\Bigl(\frac 1 3 -\frac {\alpha-\beta_j}{2\pi i}\Bigr)
\Gamma\Bigl(\frac 1 3 +\frac {\alpha-\beta_j}{2\pi i}\Bigr)(3\alpha-\sum\beta_m)e^{\frac 1 2(-\alpha+\sum\beta_m)}d\alpha\nonumber\\
&=\frac{(2\pi \Gamma(\frac 2 3))^2}{\Gamma(\frac 4 3)}
\prod_{k<j}\Gamma\Bigl(\frac 2 3 -\frac {\beta_k-\beta_j}{2\pi i}\Bigr)
\Gamma\Bigl(\frac 2 3 +\frac {\beta_k-\beta_j}{2\pi i}\Bigr)
\sum e^{\beta_m}\,,\label{1}\tag{1}
\end{align}
where $|\text{Im}~\beta_j|<2\pi/3$.
The derivation uses some non standard techniques developed by the author with the regard to integrable models in quantum field theory.

Q: Can anybody derive this integral directly?

\eqref{1} looks like formula $(3.6.6)$ in the book "Special functions" by Andrews, Askey, and Roy
\begin{align}
\frac1{2\pi i}&\int\Gamma(a+s)\Gamma(b+s)\Gamma(c+s)\Gamma(a-s)\Gamma(b-s)\Gamma(c-s)\cos\pi s\,ds\\
&=\frac{\Gamma(a)\Gamma(b)\Gamma(c)\Gamma(a+b)\Gamma(b+c)\Gamma(c+a)}{2\Gamma(a+b+c)}\label{2}\tag{2}
\end{align}
where the contour of integration is along the imaginary axis suitably deformed so that increasing sequence of poles is separated from the decreasing sequence of poles. One can see that they both contain $3$ independent parameters. I tried to show that \eqref{1} is a consequence of \eqref{2} but so far couldn't.
 A: This is a verificational proof using Liouville's theorem.
Let $C$ be a contour from $-\infty$ to $+\infty$ that separates the poles of $\Gamma\Big(\frac 1 3 -\frac {z-\beta_j}{2\pi i}\Big)$ from the poles of $\Gamma\Big(\frac 1 3 +\frac {z-\beta_j}{2\pi i}\Big)$. Consider the function
$$
f(\beta_1)=\int_C \prod_{j=1}^3\Gamma\Bigl(\frac 1 3 -\frac {z-\beta_j}{2\pi i}\Bigr)
\Gamma\Bigl(\frac 1 3 +\frac {z-\beta_j}{2\pi i}\Bigr)(3z-\sum\beta_m)e^{\frac 1 2(-z+\sum\beta_m)}dz,\tag{4}
$$
as a function of a complex variable $\beta_1$ for fixed $\beta_{2,3}\in\mathbb{C}$.
How $(4)$ is related to $(1)$? If $|\text{Im}~\beta_j|<2\pi/3$ then one can choose the contour $C$ in $(4)$ to be the real line $(-\infty,+\infty)$.
The integral $(4)$ converges absolutely, so the function $f(\beta_1)$ is meromorphic. $f(\beta_1)$ can have singularities only when two or more poles of the function 
$$
\prod_{j=1}^3\Gamma\Bigl(\frac 1 3 -\frac {z-\beta_j}{2\pi i}\Bigr)
\Gamma\Bigl(\frac 1 3 +\frac {z-\beta_j}{2\pi i}\Bigr)
$$
come close to each other. If $\beta_{2,3}$ assume generic values then only two poles can be close to each other.
1) Suppose first that the poles of $\Gamma\Big(\frac 1 3 -\frac {z-\beta_l}{2\pi i}\Big)$ and $\Gamma\Big(\frac 1 3 -\frac {z-\beta_m}{2\pi i}\Big)$ are close. Then both these poles lie to the same side of the contour. In this case one can pull the contour away from these poles. So there is not any singular contribution to $f$ from these poles.
2) However if the poles of $\Gamma\Big(\frac 1 3 -\frac {z-\beta_l}{2\pi i}\Big)$ and $\Gamma\Big(\frac 1 3 +\frac {z-\beta_m}{2\pi i}\Big)$, where $l\neq m$, are close to each other one can not pull away the contour $C$ from these poles, because the contour is squeezed between them. Such singularity gives a contribution of the order of $1/d$, where $d$ is the distance between the poles. We see that this gives a simple pole of $f$.
So let 
$$\frac 1 3 -\frac {z_1-\beta_1}{2\pi i}=-k,\quad \frac 1 3 +\frac {z_2-\beta_2}{2\pi i}=-m,$$
where $k,m$ are non-negative integers. Then
$$
z_1-z_2=2\pi i\Big(\frac23+k+m\Big)+\beta_1-\beta_2.
$$
Since according to assumption $z_1\approx z_2$ we expect a simple pole of $f$ at $\beta_1=\beta_2-2\pi i\Big(\frac23+k+m\Big)$. By symmetry all the poles of $f$ are
$$
\beta_1=\beta_{2,3}\pm2\pi i\Big(\frac23+n\Big),
$$
where $n$ is non-negative integer.
For a given $n$ there will be $n+1$ terms (such that $k+m=n$) that give singular contribution to $f$.
First, we show that the function 
$$
f(\beta_1)-\frac{(2\pi \Gamma(\frac 2 3))^2}{\Gamma(\frac 4 3)}
\prod_{k<j}\Gamma\Bigl(\frac 2 3 -\frac {\beta_k-\beta_j}{2\pi i}\Bigr)
\Gamma\Bigl(\frac 2 3 +\frac {\beta_k-\beta_j}{2\pi i}\Bigr)
\sum e^{\beta_m}\tag{5}
$$
does not have any singularities. After tedious but straightforward calculation the condition that $(5)$ does not have any poles can be written as
$$
\sum _{k=0}^n \left(b+\frac{1}{3}-n+3 k\right) \frac{ (-n)_k\left(b+\frac{2}{3}\right)_k \left(\frac{2}{3}\right)_k}{k! (b+1)_k \left(\frac{1}{3}-n\right)_k}=\left(b+\frac{1}{3}\right)\frac{\left(\frac{4}{3}\right)_n \left(b+\frac{4}{3}\right)_n}{\left(\frac{2}{3}\right)_n (b+1)_n}.\tag{3}
$$
$(3)$ becomes a polynomial identity after multiplication of both sides with $(b+1)_n$. Here we used the notation for the Pochhammer symbol $(x)_n=x(x+1) ... (x+n-1)$. This sum can be proved using contiguous relation http://dlmf.nist.gov/16.4.E12 and Pfaff-Saalschutz summation formula.
So $(5)$ is an entire function. One can show that $(5)$ decays exponentially when $|\text{Re}~\beta_1|\to \infty$ and grows at most as $|\text{Im}~\beta_1|^{2/3}$ for large $|\text{Im}~\beta_1|$. It follows according to generalized Liouville's theorem that $(5)$ is constant. The value of the constant is $0$ as can be seen by considering asymptotics when  when $|\text{Re}~\beta_1|\to \infty$.
