Evaluate the Integral using Contour Integration (Theorem of Residues) $$
J(a,b)=\int_{0}^{\infty }\frac{\sin(b x)}{\sinh(a x)} dx
$$
This integral is difficult because contour integrals normally cannot be solved with a sin(x) term in the numerator because of singularity issues between the 1st quadrant to the 2nd quadrant in the upper half circle of the contour.  
I've assumed:
$$
\sin(bx)=e^{ibz}
$$
since it effectively IS the sine term in the upper half circle.  
I've also used the following substitutions:
$$
x=z  ;  z=re^{i\theta}   ;  z=\cos(\theta)+i\sin(\theta)
$$
From what I understand, the x range has to include all values somehow.
When I plug in the substitutions and the Euler forms of sin and sinh, the integral becomes:
$$
2\int_{-\infty }^{\infty }\frac{e^{ibx}}{e^{ax}-e^{ax}}dx
$$
Have I screwed up the subs?  Did I reduce incorrectly?  Not sure what to go from here. If anyone could help shed some light it would be very much appreciated.  

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\begin{align}
\on{J}\pars{a,b} & \equiv
\bbox[5px,#ffd]{\int_{0}^{\infty }{\sin\pars{bx} \over \sinh\pars{ax}}\,\dd x} =
\on{sgn}\pars{a}\int_{0}^{\infty }{\sin\pars{bx} \over \sinh\pars{\verts{a}x}}\,\dd x
\\[5mm] & \stackrel{x\ =\ t/\pars{2\verts{a}}}{=}\,\,\,
\left.{1 \over 2a}\int_{0}^{\infty }{\sin\pars{\beta t/2} \over \sinh\pars{t/2}}\,\dd t
\,\right\vert_{\,\ds{\beta\ \equiv\ b/\verts{a}}}
\\[5mm] & =
-\,{\ic \over 2a}\int_{0}^{\infty }
{\expo{-\pars{1 - \beta\,\ic}t/2}\,\,\, -\,
\expo{-\pars{1 + \beta\,\ic}t/2} \over 1 - \expo{-t}}\,\dd t
\\[5mm] & =
-\,{\ic \over 2a}\left[%
\int_{0}^{\infty }
{1 - \expo{-\pars{1 + \beta\,\ic}t/2} \over 1 - \expo{-t}}
\,\dd t\right.
\\[2mm] & \phantom{-\,{\ic \over 2a}\,\,\,\,\,}\left.
-\int_{0}^{\infty }
{1 - \expo{-\pars{1 - \beta\,\ic}t/2} \over 1 - \expo{-t}}
\,\dd t\right]\label{1}\tag{1}
\\[5mm] & =
-\,{\ic \over 2a}\bracks{%
\Psi\pars{{1 \over 2} + {\beta\,\ic \over 2}} -
\Psi\pars{{1 \over 2} - {\beta\,\ic \over 2}}}
\label{2}\tag{2}
\\[5mm] & =
-\,{\ic \over 2a}\braces{%
\pi\cot\pars{\pi\bracks{%
{1 \over 2} - {\beta\,\ic \over 2}}}} =
-\,{\pi\ic \over 2a}\tan\pars{\beta\,\ic \over 2}
\label{3}\tag{3}
\\[5mm] & =
\bbx{{\pi \over 2a}\tanh\pars{b \over 2\verts{a}}} \\ &
\end{align}

*

*$\ds{\Psi}$: Digamma Function.

*(\ref{2}): See $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table.

*(\ref{3}): Euler Reflection Formula. See $\ds{\color{black}{\bf 6.3.7}}$ in A & S Table.

