Using a rectangular contour in the complex plane, bypassing the poles at $z=0$ and $z=i$, i got $$\int_{-\infty}^{+\infty}\frac{\sinh(ax)e^{ikx}}{\sinh(\pi x)}dx - e^{-k}\int_{-\infty}^{+\infty}\frac{\sinh[a(x+i)]e^{ikx}}{\sinh(\pi x)}dx = \sin(a)e^{-k}$$
and now?
On
This is not exactly straightforward. We are going to have to work out two contour integrals to get what we need here. To see this, let's consider the contour integral that comes naturally:
$$\oint_C dz \frac{\sinh{a z}}{\sinh{\pi z}} e^{i k z} $$
where $C$ is the rectangle having vertices $-R$, $R$, $R+i$, $-R+i$, with a small semicircular indent of radius $\epsilon$ into the rectangle at $z=i$. (We have to introduce this indent because the integral along the line $\operatorname{Im}{z}=1$ does not converge.) The contour integral is then
$$\int_{-R}^R dx \frac{\sinh{a x}}{\sinh{\pi x}} e^{i k x} + i \int_0^1 dy \, \frac{\sinh{a (R+i y)}}{\sinh{\pi (R+i y)}} e^{i k (R+i y)} \\ + PV \int_{R}^{-R} dx \frac{\sinh{a (x+i)}}{\sinh{\pi (x+i)}} e^{i k (x+i)} + i \int_1^0 dy \, \frac{\sinh{a (-R+i y)}}{\sinh{\pi (-R+i y)}} e^{i k (-R+i y)}\\ + i \epsilon \int_{2 \pi}^{\pi} d\theta \, e^{i \theta} \frac{\sinh{\left [a \left (i+ \epsilon e^{i \theta}\right ) \right ]}}{\sinh{\left [\pi \left (i+ \epsilon e^{i \theta}\right ) \right ]}} e^{i k (i+\epsilon e^{i \theta})}$$
where $PV$ denotes a Cauchy principal value. Because $|a| \lt \pi$, the second and fourth integrals vanish as $R \to \infty$. Note also, as $\epsilon \to 0$, the fifth integral approaches
$$i \epsilon \int_{2 \pi}^{\pi} d\theta \, e^{i \theta} \frac{\sinh{(i a)}}{-\pi \epsilon e^{i \theta}} e^{-k} = -e^{-k} \sin{a}$$
The contour integral is equal to zero by Cauchy's Theorem. Thus we have
$$\left (1+e^{-k} \cos{a} \right ) \int_{-\infty}^{\infty} dx \frac{\sinh{a x}}{\sinh{\pi x}} e^{i k x} +i e^{-k} \sin{a}\; PV \int_{-\infty}^{\infty} dx \frac{\cosh{a x}}{\sinh{\pi x}} e^{i k x} = e^{-k} \sin{a}$$
Unfortunately, we have two unknown integrals and only one equation. Thus, it makes sense to consider
$$\oint_C dz \frac{\cosh{a z}}{\sinh{\pi z}} e^{i k z} $$
where $C$ now has an extra semicircular detour into the rectangle at $z=0$. The contour integral is then
$$\int_{-R}^R dx \frac{\cosh{a x}}{\sinh{\pi x}} e^{i k x} + i \int_0^1 dy \, \frac{\cosh{a (R+i y)}}{\sinh{\pi (R+i y)}} e^{i k (R+i y)} \\ + \int_{R}^{-R} dx \frac{\cosh{a (x+i)}}{\sinh{\pi (x+i)}} e^{i k (x+i)} + i \int_1^0 dy \, \frac{\cosh{a (-R+i y)}}{\sinh{\pi (-R+i y)}} e^{i k (-R+i y)}\\ + i \epsilon \int_{2 \pi}^{\pi} d\theta \, e^{i \theta} \frac{\cosh{\left [a \left (i+ \epsilon e^{i \theta}\right ) \right ]}}{\sinh{\left [\pi \left (i+ \epsilon e^{i \theta}\right ) \right ]}} e^{i k (i+\epsilon e^{i \theta})} + i \epsilon \int_{\pi}^{0} d\theta \, e^{i \theta} \frac{\cosh{\left [a \left (\epsilon e^{i \theta}\right ) \right ]}}{\sinh{\left [\pi \left (\epsilon e^{i \theta}\right ) \right ]}} e^{i k (\epsilon e^{i \theta})}$$
Using a similar analysis as above, i.e., Cauchy's theorem and taking the limit as $R \to \infty$ and $\epsilon \to 0$, we find that
$$ i e^{-k} \sin{a}\int_{-\infty}^{\infty} dx \frac{\sinh{a x}}{\sinh{\pi x}} e^{i k x} + \left (1+e^{-k} \cos{a} \right )\; PV \int_{-\infty}^{\infty} dx \frac{\cosh{a x}}{\sinh{\pi x}} e^{i k x} = i \left (1-e^{-k} \cos{a} \right )$$
Let the original integral be $A$ and the principal value integral we picked up along the way be $B$. We now have two equations and two unknowns:
$$\left (1+e^{-k} \cos{a} \right ) A + i e^{-k} \sin{a}\; B = e^{-k} \sin{a}$$ $$i e^{-k} \sin{a}\; A + \left (1+e^{-k} \cos{a} \right ) B = i \left (1-e^{-k} \cos{a} \right )$$
We may then solve this system with relative ease. Sparing the reader some mild algebra, I find the desired result for $A$:
$$A = \int_{-\infty}^{\infty} dx \frac{\sinh{a x}}{\sinh{\pi x}} e^{i k x} = \frac{\sin{a}}{\cosh{k} + \cos{a}} $$
ADDENDUM
May as well look at the extra integral we get for free.
$$B = PV \int_{-\infty}^{\infty} dx \frac{\cosh{a x}}{\sinh{\pi x}} e^{i k x} = i \frac{\sinh{k}}{\cosh{k} + \cos{a}} $$
What's interesting is that only the imaginary part of the integral is nonzero. In this case, the principal value s not necessary, and we get that
$$\int_{-\infty}^{\infty} dx \frac{\cosh{a x}}{\sinh{\pi x}} \sin{k x} = \frac{\sinh{k}}{\cosh{k} + \cos{a}} $$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty} {\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x \,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}} = 2\,\Re\int_{0}^{\infty} {\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x \\[5mm] \stackrel{2\pi x\ \mapsto\ x}{=}\,\,\, &\ {1 \over \pi}\,\Re\int_{0}^{\infty} {\sinh\pars{\alpha x}\expo{\ic \kappa x} \over \sinh\pars{x/2}}\,\dd x\quad\mbox{where}\quad \left\{\begin{array}{rcl} \ds{\alpha} & \ds{\equiv} & \ds{a \over 2\pi} \\[1mm] \ds{\kappa} & \ds{\equiv} & \ds{k \over 2\pi} \end{array}\right. \end{align} Then, \begin{align} &\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty} {\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x \,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}} \\[5mm] = &\ {1 \over \pi}\,\Re\int_{0}^{\infty} {\expo{-\pars{-\alpha - \ic\kappa}x} - \expo{-\pars{\alpha - \ic\kappa}x} \phantom{A}\over \expo{x/2} - \expo{-x/2}}\,\dd x \\[5mm] = &\ {1 \over \pi}\,\Re\int_{0}^{\infty} {\expo{-\pars{-\alpha - \ic\kappa + 1/2}x}\,\,\, - \expo{-\pars{\alpha - \ic\kappa + 1/2}x} \phantom{A}\over 1 - \expo{-x}}\,\dd x \\[5mm] = &\ {1 \over \pi}\,\Re\int_{0}^{\infty} {\expo{-x} - \expo{-\pars{\alpha - \ic\kappa + 1/2}x} \phantom{A}\over 1 - \expo{-x}}\,\dd x \\[2mm] - &\ {1 \over \pi}\,\Re\int_{0}^{\infty} {\expo{-x} - \expo{-\pars{-\alpha - \ic\kappa + 1/2}x} \phantom{A}\over 1 - \expo{-x}}\,\dd x \\[5mm] = &\ {1 \over \pi}\,\Re\bracks{% \Psi\pars{{1 \over 2} + \alpha - \ic\kappa} - \Psi\pars{{1 \over 2} - \alpha - \ic\kappa}} \end{align} See $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table
In addition, \begin{align} &\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty} {\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x \,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}} \\[5mm] = &\ {1 \over 2\pi}\bracks{% \Psi\pars{{1 \over 2} + \alpha - \ic\kappa} - \Psi\pars{{1 \over 2} - \alpha + \ic\kappa}} \\[2mm] - &\ {1 \over 2\pi}\bracks{% \Psi\pars{{1 \over 2} - \alpha - \ic\kappa} - \Psi\pars{{1 \over 2} + \alpha + \ic\kappa}} \end{align}
With Euler Reflection Formula ( see ${\color{black}{\bf 6.3.7}}$ in A & S Table ): \begin{align} &\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty} {\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x \,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}} \\[5mm] = &\ {1 \over 2}\braces{% \cot\pars{\pi\bracks{{1 \over 2} - \alpha + \ic\kappa}} - \cot\pars{\pi\bracks{{1 \over 2} + \alpha + \ic\kappa}}} \\[5mm] = &\ {1 \over 2}\bracks{% \tan\pars{{a \over 2} - \ic\,{k \over 2}} + \tan\pars{{a \over 2} + \ic\,{k \over 2}}} \\[5mm] = &\ \bbx{\sin\pars{a} \over \cosh\pars{k} + \cos\pars{a}} \\ & \end{align}