Singularities of an integral We have the integral :
$$I(t)=-i\int_0^\infty \frac{\log\left[\frac{\sin(t\log\sqrt{1+ix})}{\log(1+ix)} \right ]-\log\left[\frac{\sin(t\log\sqrt{1-ix})}{\log(1-ix)} \right ]}{e^{2\pi x}-1} \, dx$$
I have tried everything to compute the integral, but it seems it's not doable in terms of elementary functions. For instance, the form of integral suggests that the Able-Plana formula can be used, but it can't. And closing the contour is troublesome. I have reasons to believe that the integral has logarithmic singularities, and can be expressed as:
$$I(t)=f(t)+\sum_{\beta_{j}}\log\left(1-\frac{t^{2}}{\beta_{j}^{2}}\right)$$
Where $f(t)$ is an even,entire function -possibly zero- and the numbers $\beta_{j}$ are positive, real numbers. However, i haven't been able to prove that. A plot of the function (numerical integration) could be helpful.
EDIT 
We can express the integral as :
$$\int_{1-i\infty}^{1+i\infty}\frac{\log \left[\frac{ \sin{\left(\frac{t}{2}\log{u}\right)}}{\log{u}} \right ]}{e^{2\pi i u}-1}du-\int_{1}^{1+i\infty}\log \left[\frac{ \sin{\left(\frac{t}{2}\log{u}\right)}}{\log{u}} \right ]du$$
EDIT 2
The derivative of the integral can be expressed as :
$$\frac{d}{dt}I(t)=\frac{t}{2i}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{\frac{\log(1+ix)^{2}}{\left(\frac{t}{2}\log(1+ix) \right )^{2}-\pi ^{2}n^{2}} -\frac{\log(1-ix)^{2}}{\left(\frac{t}{2}\log(1-ix) \right )^{2}-\pi ^{2}n^{2}} }{e^{2\pi x}-1}$$
We can do the integral if we can calculate 
$$\int_{0}^{\infty}\frac{\log(1\pm ix)^{2}}{\left(\frac{t}{2}\log(1 \pm ix) \right )^{2}-\pi ^{2}n^{2}}e^{-2\pi mx}dx$$
 A: As discussion
This is seem to me is of Bromwich type integral.
\begin{align}
 I\left( t \right) &= \int_0^\infty  {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right) - \log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 - ix} } \right)} \over {\log \left( {1 - ix} \right)}}}} \right)}}{{e^{2\pi x}  - 1}}dx}  \\ 
  &= \int_0^\infty  {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x}  - 1}}dx}  - \int_0^\infty  {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 - ix} } \right)} \over {\log \left( {1 - ix} \right)}}}} \right)}}{{e^{2\pi x}  - 1}}dx}  \\ 
  &= \int_0^\infty  {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x}  - 1}}dx}  - \int_0^{ - \infty } {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)}}{{e^{ - 2\pi y}  - 1}}d\left( { - y} \right)}  \\ 
  &= \int_0^\infty  {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x}  - 1}}dx}  - \int_{ - \infty }^0 {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)}}{{e^{ - 2\pi y}  - 1}}dy}  \\ 
  &= \int_0^\infty  {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x}  - 1}}dx}  + \int_{ - \infty }^0 {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)}}{{1 - e^{ - 2\pi y} }}dy}  
 \end{align}
where we used the substitution  $-x=y$ in the second integral.
Now, using geometric series we have
\begin{align}
\frac{1}{{e^{2\pi x}  - 1}} = \frac{{e^{ - 2\pi x} }}{{1 - e^{ - 2\pi x} }} = e^{ - 2\pi x} \sum\limits_{n = 0}^\infty  {e^{ - 2n\pi x} }  = \sum\limits_{n = 0}^\infty  {e^{ - 2\left( {n + 1} \right)\pi x} } 
\end{align}
so that 
\begin{align}
\int_0^\infty  {\left( {\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)} \right)\sum\limits_{n = 0}^\infty  {e^{ - 2\left( {n + 1} \right)\pi x} } dx}  + \int_{ - \infty }^0 {\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)\sum\limits_{n = 0}^\infty  {e^{ - 2n\pi y} } dy} 
\end{align}
Let $ u = \sqrt {1 + iy}  \Rightarrow u^2  = 1 + iy \Rightarrow 2udu = idy \Rightarrow dy =  - 2iudu $
\begin{align} 
 I(t):= - 2i\int_1^{1 + i\infty } {\left( {\log \left( {{\textstyle{{\sin \left( {t\log u } \right)} \over {\log \left( {u^2} \right)}}}} \right)} \right)\sum\limits_{n = 0}^\infty  {e^{  2i\left( {n + 1} \right)\pi 
  \left( {u^2  - 1} \right)
} } udu}  
\\
&- 2i\int_{1 - i\infty }^1 {\log \left( {{\textstyle{{\sin \left( {t\log u } \right)} \over {\log \left( {u^2} \right)}}}} \right)\sum\limits_{n = 0}^\infty  {e^{  2in\pi 
 \left( {u^2  - 1} \right)
} } udu}  
\end{align}
Now use the fact that:
Theorem:
Let $F$ be an analytic dunction whose singularities $z_1,z_2,\cdots,z_n$ belong to the half-plane $\{{z|\Re({z})<c}\}$ and let $
\mathop {\lim }\limits_{z \to \infty } F\left( z \right) = 0$. Then
\begin{align}
\frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{zt} F\left( z \right)dz}  = \sum\limits_{k = 1}^n {\mathop {{\mathop{\rm Res}\nolimits} }\limits_{z = z_k } \left\{ {e^{zt} F\left( z \right)} \right\}} 
\end{align}
