Help with a troublesome double integral I'm having difficulty with a double integral

$$-2i\int_{0}^{\infty}\int_{0}^{\infty}\frac{dxdt}{t(e^{2\pi x}-1)(e^{2\pi t/s}-1)}\left[\cos(t\log(1-ix))-\cos(t\log(1+ix))\right]$$

where $s\in\text{D}\subset \mathbb{C}$, but i am not sure of the domain of convergence !
EDIT
It can be shown that the above integral is equivalent to :
$$-4\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sinh\left(t\tan^{-1}(x)\right )\sin\left(t\log\left(\sqrt{1+x^{2}} \right ) \right )}{t(e^{2\pi x}-1)(e^{2\pi t/s}-1)}dxdt$$
EDIT2
Doing the integral with respect to $t$ first appears to be a promising approach. Particularly, we wish to compute the integral of the form:
$$\int_{0}^{\infty}\frac{\sinh\left(at\right )\sin\left(bt \right )}{t(e^{2\pi tz}-1)}dt\;\;\;\;\;\left(a,b\in \mathbb{R}^{ \geq  0}\;\;,\;\;z\in\text{D}\subset \mathbb{C}\right)$$
 A: The integral to evaluate is
$$I(s):=-2i\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos{\left(t\log{(1-ix)}\right)}-\cos{\left(t\log{(1+ix)}\right)}}{t\left(e^{2\pi x}-1\right)\left(e^{2\pi t/s}-1\right)}\mathrm{d}x\mathrm{d}t,$$
where $s\in\mathbb{C}$ is a complex parameter. Right away we notice that the integration with respect to the variable $x$ is much more formidable than that with respect to $t$, so the first thing we do is interchange the order of integration:
$$\begin{align}I(s)&=-2i\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos{\left(t\log{(1-ix)}\right)}-\cos{\left(t\log{(1+ix)}\right)}}{t\left(e^{2\pi x}-1\right)\left(e^{2\pi t/s}-1\right)}\mathrm{d}t\mathrm{d}x\\
&=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{0}^{\infty}\frac{\cos{\left(t\log{(1-ix)}\right)}-\cos{\left(t\log{(1+ix)}\right)}}{t\left(e^{2\pi t/s}-1\right)}\mathrm{d}t.\end{align}$$
The integration w.r.t. $t$ is still too complicated to evaluate, but notice that the term in the numerator of the integrand involving a difference of cosine functions is highly suggestive of the Fundamental Theorem of Calculus. Indeed, using the simple identity $\int_{a}^{b}\sin{(t\,\omega)}\,\mathrm{d}\omega=\frac{\cos{(t\,a)}-\cos{(t\,b)}}{t}$ we can write,
$$\int_{\log{(1-ix)}}^{\log{(1+ix)}}\sin{(t\,\omega)}\,\mathrm{d}\omega=\frac{\cos{(t\,\log{(1-ix)})}-\cos{(t\,\log{(1+ix)})}}{t}.$$
Substituting this expression back into the integral over $t$ not only absorbs the problematic factor of $t$ in the denominator of the integrand, it also gives us the option of performing another change-of-order-of-integration magic trick.
$$\begin{align}I(s)&=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{0}^{\infty}\frac{\mathrm{d}t}{\left(e^{2\pi t/s}-1\right)}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\sin{(t\,\omega)}\,\mathrm{d}\omega\\
&=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{0}^{\infty}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\frac{\sin{(t\,\omega)}}{\left(e^{2\pi t/s}-1\right)}\,\mathrm{d}\omega\mathrm{d}t\\
&=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{\left(e^{2\pi t/s}-1\right)}\,\mathrm{d}t\mathrm{d}\omega\end{align}$$
So we turn our attention to solving the integral $\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{e^{2\pi t/s}-1}\mathrm{d}t$. This integral (see notes at bottom) is,
$$\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{e^{2\pi t/s}-1}\mathrm{d}t=\frac{1}{2\omega}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right),$$
and the integral becomes,
$$\begin{align}I(s)&=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{2\omega}\\
&=-i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)} \int_{\log{(1-ix)}}^{\log{(1+ix)}} \left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{\omega}\\
&=-i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)} G(x,s).\end{align}$$
See appendix 2 for details on the function $G(x,s)$. It is seen to have the form $G(x,s)=f(ix)-f(-ix)$, and so we can apply the Abel-Plana formula to the final integral $I(s)$:
$$\begin{align}
I(s)&=-i\int_{0}^{\infty}\frac{f(ix)-f(-ix)}{\left(e^{2\pi x}-1\right)}\mathrm{d}x\\
&=\int_{0}^{\infty}f(x)\,\mathrm{d}x+\frac12f(0)-\sum_{n=0}^{\infty}f(n)
\end{align}$$

Appendix 1:
$\tau=\frac{2\pi t}{s}$, $t=\frac{s}{2\pi}\tau$, $\alpha:=\frac{s\omega}{2\pi}$
$$\begin{align}\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{e^{2\pi t/s}-1}\mathrm{d}t&=\frac{s}{2\pi}\int_{0}^{\infty}\frac{\sin{(\omega\frac{s}{2\pi}\tau)}}{e^{\tau}-1}\mathrm{d}\tau\\
&=\frac{s}{2\pi}\int_{0}^{\infty}\frac{\sin{(\alpha\,\tau)}}{e^{\tau}-1}\mathrm{d}\tau\\
&=\frac{s}{2\pi}\int_{0}^{\infty}\frac{\sin{(\alpha\,\tau)}\,e^{-\tau}}{1-e^{-\tau}}\mathrm{d}\tau\\
&=\frac{s}{2\pi}\int_{0}^{\infty}\sin{(\alpha\,\tau)}\,e^{-\tau}\sum_{n=0}^{\infty}e^{-n\tau}\mathrm{d}\tau\\
&=\frac{s}{2\pi}\sum_{n=0}^{\infty}\int_{0}^{\infty}\sin{(\alpha\,\tau)}\,e^{-(n+1)\tau}\mathrm{d}\tau\\
&=\frac{s}{2\pi}\sum_{n=0}^{\infty}\frac{\alpha}{\alpha^2+(n+1)^2}\\
&=\frac{s}{2\pi}\frac{\pi\alpha\coth{(\pi\alpha)}-1}{2\alpha}\\
&=\frac{1}{2\omega}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\end{align}.$$

Appendix 2:
$$\int\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{\omega}=\log{\left(\sinh{\left(\frac{s\omega}{2}\right)}\right)}-\log{\omega}+\text{constant}$$
$$\begin{align}
G(x,s):&=\int_{\log{(1-ix)}}^{\log{(1+ix)}}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{\omega}\\
&=\left(\log{\left(\sinh{\left(\frac{s\log{(1+ix)}}{2}\right)}\right)}-\log{\log{(1+ix)}}\right)-\left(\log{\left(\sinh{\left(\frac{s\log{(1-ix)}}{2}\right)}\right)}-\log{\log{(1-ix)}}\right)\\
&=f(ix)-f(-ix),
\end{align}$$
where $f(z):=\left(\log{\left(\sinh{\left(\frac{s\log{(1+z)}}{2}\right)}\right)}-\log{\log{(1+z)}}\right)$.

In response to OP's edit #2:
For $\Re{(\gamma)}>|\Re{(\beta)}|$,
$$\int_{0}^{\infty}\frac{\cos{\left(\alpha\,t\right)}\sinh{\left(\beta\,t\right)}}{e^{\gamma\,t}-1}\,\mathrm{d}t = \frac{\beta}{2\left(\alpha^2+\beta^2\right)}-\frac{\pi}{2\gamma}\cdot\frac{\sin{\left(\frac{2\pi\beta}{\gamma}\right)}}{\cosh{\left(\frac{2\pi\alpha}{\gamma}\right)}-\cos{\left(\frac{2\pi\beta}{\gamma}\right)}}.$$
The above integral is formula $4.132.4$ of Gradstheyn's Table of integrals.
