Evaluate $\int_0^{\infty } \Bigl( 2qe^{-x}-\frac{\sinh (q x)}{\sinh \left(\frac{x}{2}\right)} \Bigr) \frac{dx}x$ Gradshteyn&Ryzhik $3.554.5$ states that:
$$\int_0^{\infty } \frac1x \biggl( 2qe^{-x}-\frac{\sinh (q x)}{\sinh \left(\frac{x}{2}\right)} \biggr) \, dx=\log\bigl(\cos (\pi  q) \bigr)+2 \log \left(\Gamma \bigl(q+\frac12 \bigr)\right)-\log (\pi )~, \quad q^2<\frac{1}{4}$$
It seems like to have sth to do with the Binet  representation of log-gamma function, but I haven't figured out how to solve it. Any kind of help will be appreciated.
 A: The formula of the question follows easily from Ramanujan's generalization of Frullani's integral.  See 'The Quarterly Reports of S. Ramanujan,' American Mathematical Monthly, Vol. 90, #8 Oct 1983, p 505-516.  I won't give the conditions, but state it to establish the notation.  Let
$$ f(x)-f(\infty)=\sum_{k=0}^\infty u(k)(-x)^k/k! \quad,\quad g(x)-g(\infty)=\sum_{k=0}^\infty v(k)(-x)^k/k!$$
$$ f(0)=g(0) \quad,\quad f(\infty)=g(\infty) $$
Then
$$\int_0^\infty \frac{dx}{x} \big(f(ax) - g(bx) \big)= \big(f(0)-f(\infty) \big)\Big( \log{(b/a)} + 
\frac{d}{ds} \log{\Big(\frac{v(s)}{u(s)}\Big)}\Big|_{s=0} \Big) $$
For the OP's case, a=b=1, $f(x)=2qe^{-x} \implies u(k)=2q, \ f(0)=2q, f(\infty)=0.$
We will eventually show
$$ (1) \quad g(x)=\frac{\sinh(q \ x)}{\sinh(x/2)} = -\sum_{n=0}^\infty \frac{(-x)^n}{n!} \ \cos(\pi \ n) \big( \zeta(-n, 1/2+q) - \zeta(-n, 1/2-q) \big)$$
where $\zeta(s,a)$ is the Hurwitz zeta function.
Given (1), it is easy to see that
$$ \frac{d}{ds} \log{v(s)} \big|_{s=0} = \frac{v'(0)}{v(0)} = -\frac{ \zeta'(0, 1/2+q)-\zeta'(0, 1/2-q)}{ \zeta(0, 1/2+q)-\zeta(0, 1/2-q) }$$
However, it is known that
$$\zeta'(0,a)=\log(\Gamma(a)/\sqrt{2\pi}) \text{ and  } \zeta(0,a)=-B_1(a)=1/2-a $$ where in the last formula the Hurwitz zeta has been connected to the Bernoulli polynomial by the formula
$$ (2) \quad \zeta(-n,a) = -\frac{B_{n+1}(a)}{n+1}. $$
Doing the rest of the algebra results in the expression 
$$ (3) \quad \int_0^\infty \Big(2qe^{-x} - \frac{ \sinh(q \ x)}{\sinh{x/2} } \Big) \frac{dx}{x} = \log{\Gamma(1/2+q)} - \log{\Gamma(1/2-q)}. $$  To get it in the form of the OP's request, use the gamma-function reflection formula 
$$ \Gamma(1/2-q)\Gamma(1/2+q) = \frac{\pi}{\cos{\pi q } } .$$
Now, to prove (1):
$$ \frac{\sinh(q \ x)}{\sinh(x/2)} = \frac{e^{qx} - q^{-qx}}{e^{x/2}-e^{-x/2}} =
\frac{1}{x}\Big( \frac{x}{e^x-1} \exp(x(1/2+q))+\frac{x}{e^x-1} \exp(x(1/2-q)) \Big) 
$$
Use the well-known generating function for Bernoulli polynomials,
$$ \frac{\sinh(q \ x)}{\sinh(x/2)} =\frac{1}{x}\sum_{n=0}^\infty \frac{x^n}{n!} \Big( B_n(1/2+q) - B_n(1/2-q) \Big) =\sum_{n=0}^\infty \frac{x^n}{n!(n+1)} \Big( B_{n+1}(1/2+q) - B_{n+1}(1/2-q) \Big)  $$
where in the second step we have re-indexed because $B_0(x)=1$ and the first term is thus zero.  Then use (2) in the last formula to complete the proof of (1).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{q \in \mathbb{R}}$, note that
\begin{align}
&\bbox[10px,#ffd]{\left.\int_{0}^{\infty}{1 \over x}\bracks{2q\expo{-x} -
{\sinh\pars{qx} \over \sinh\pars{x/2}}}\dd x
\,\right\vert_{{\large q\ \in\ \mathbb{R}} \atop
{\large q^{2}\ <\ 1/4}}}
\\[5mm] = &\
\left.\mrm{sgn}\pars{q}\int_{0}^{\infty}{1 \over x}
\bracks{2\verts{q}\expo{-x} -
{\sinh\pars{\verts{q}x} \over \sinh\pars{x/2}}}\dd x
\,\right\vert_{\ \verts{q}\ <\ 1/2}
\label{1}\tag{1}
\end{align}
The last integral becomes:
\begin{align}
&\!\!\!\!\!\!\!\!\!\!\bbox[10px,#ffd]{\int_{0}^{\infty}{1 \over x}\bracks{2\verts{q}\expo{-x} -
{\expo{-\pars{1/2 - \verts{q}}x} - \expo{-\pars{1/2 + \verts{q}}x} \over 1 - \expo{-x}}}\dd x}
\\[5mm] \stackrel{x\ =\ -\ln\pars{t}}{=} &\
\int_{1}^{0}{1 \over -\ln\pars{t}}\pars{2\verts{q}t -
{t^{1/2 - \verts{q}} -  t^{1/2 + \verts{q}} \over 1 - t}}
\pars{-\,{\dd t \over t}}
\\[5mm] = &\  \!\!\!\!
\int_{0}^{1}\!\!\underbrace{\bracks{-\,{1 \over \ln\pars{t}}}}
_{\ds{\int_{1}^{\infty}t^{\xi - 1}\,\dd\xi}}
{2\verts{q} - 2\verts{q}t - t^{-1/2 - \verts{q}} +  t^{-1/2 + \verts{q}}
\over 1 - t}\,
\dd t
\\[5mm] = &\!\!\!
\int_{1}^{\infty}\!\!\!\!\int_{0}^{1}\!\!
{2\verts{q}t^{\xi - 1} - 2\verts{q}t^{\xi} - t^{\xi - 3/2 - \verts{q}} +  t^{\xi - 3/2 + \verts{q}} \over 1 - t}\,\dd t\,\dd\xi
\\[5mm] = &\
\int_{1}^{\infty}\int_{0}^{1}\left[%
2\verts{q}\int_{0}^{1}{1 - t^{\xi} \over 1 - t}\,\dd t -
2\verts{q}\int_{0}^{1}{1 - t^{\xi - 1} \over 1 - t}\,\dd t\right.
\\[2mm] &\ \left. +\int_{0}^{1}{1 - t^{\xi - 3/2 - \verts{q}} \over 1 - t}\,\dd t -
\int_{0}^{1}{1 - t^{\xi - 3/2 + \verts{q}} \over 1 - t}\,\dd t
\right]\dd\xi
\\[5mm] = &\
\int_{1}^{\infty}\left[\vphantom{\huge A}\,%
2\verts{q}\Psi\pars{\xi + 1} - 2\verts{q}\Psi\pars{\xi}\right.
\\[2mm] & \phantom{\int_{1}^{\infty}\left[\right.}
\left. +\ \Psi\pars{\xi - {1 \over 2} - \verts{q}} -
\Psi\pars{\xi - {1 \over 2} + \verts{q}}\right]\dd\xi
\label{2}\tag{2}
\\[5mm] = &
\overbrace{\lim_{\xi \to \infty}\bracks{
2\verts{q}\ln\pars{\xi} + \ln\pars{\Gamma\pars{\xi - 1/2 - \verts{q}} \over \Gamma\pars{\xi - 1/2 + \verts{q}}}}}
^{\ds{=\ 0}}
\\[2mm] &\
+\ln\pars{\Gamma\pars{1/2 + \verts{q}} \over \Gamma\pars{1/2 - \verts{q}}}
\\[5mm] = &\
\ln\pars{\Gamma^{\, 2}\pars{1/2 + \verts{q}} \over
\Gamma\pars{1/2 + \verts{q}}\Gamma\pars{1/2 - \verts{q}}}
\\[5mm] = &\
\ln\pars{\Gamma^{\, 2}\pars{1/2 + \verts{q}} \over
\pi/\sin\pars{\pi\bracks{1/2 + q}}}\label{3}\tag{3}
\\[5mm] = &\
\bbx{\ln\pars{\cos\pars{\pi\verts{q}}} +
2\ln\pars{\Gamma\pars{\verts{q} + {1 \over 2}}}
- \ln\pars{\pi}}\label{4}\tag{4} \\ &
\end{align}
(\ref{2}): I used an integral representation
( see $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table ) of the digamma $\ds{\mbox{function}\ \Psi}$.

(\ref{3}): Euler Reflection Formula. See
$\ds{\color{black}{\bf 6.1.17}}$ in A & S Table.

Finally, with (\ref{1}) and (\ref{4}):
\begin{align}
&\bbox[10px,#ffd]{\left.\int_{0}^{\infty}{1 \over x}\bracks{2q\expo{-x} -
{\sinh\pars{qx} \over \sinh\pars{x/2}}}\dd x
\,\right\vert_{{\large q\ \in\ \mathbb{R}} \atop
{\large q^{2}\ <\ 1/4}}}
\\[5mm] = &\
\bbox[25px,#ffd,border:1px groove navy]{\mrm{sgn}\pars{q}\bracks{\ln\pars{\cos\pars{\pi\verts{q}}} +
2\ln\pars{\Gamma\pars{\verts{q} + {1 \over 2}}}
- \ln\pars{\pi}}}
\\ &\
\end{align}
