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.  

You can change the integration $\int_{0}^{\infty}$ to $\frac{1}{2}\int_{-\infty}^{\infty}$ . Since $\cos$ part will disappear, 
Rewrite the integrand as
$$
\frac{e^{ibx}}{e^{ax}-e^{-ax}}=\frac{\exp\{(a+ib)x\}}{\exp(2ax)-1}
$$
Take the contour $[-R,R]$, $[R,R+\frac{i\pi}{a}]$, $[R+\frac{i\pi}{a},-R+\frac{i\pi}{a}]$, $[-R+\frac{i\pi}{a}, -R]$. (We will take $R\rightarrow\infty$). 
However the integrand has singularities at $x=0$ and $x=\frac{i\pi}{a}$, so detour those points by taking upper half circle of small radius $\epsilon$ at $x=0$, and lower half circle of small radius $\epsilon$ at $x=\frac{i\pi}{a}$. Then the contour does not have singularities inside. (We will take $\epsilon\rightarrow 0$). 
Note that the integrals on the vertical sides and the small part of the horizontal lines around $0$ and $\frac{i\pi}{a}$  will vanish with $R\rightarrow\infty$ and $\epsilon\rightarrow 0$.  
Denote the integral by $I=\int_{-\infty}^{\infty}\frac{e^{ibx}}{e^{ax}-e^{-ax}}dx$. Then by Residue Theorem, we have
$$
I-I\exp\{(a+ib)\frac{i\pi}{a}\}-\frac{1}{2}2\pi i \frac{1}{2a}-\frac{1}{2}2\pi i\frac{1}{2a}\exp\{(a+ib)\frac{i\pi}{a}\}=0
$$
We solve this for $I$, it now follows that 
$$
J(a,b)=\int_{0}^{\infty }\frac{\sin(b x)}{\sinh(a x)} dx=-iI=\frac{\pi}{2a}\frac{1-\exp(-\frac{b}{a}\pi)}{1+\exp(-\frac{b}{a}\pi)}
$$
and this RHS is the integral that we wanted in the beginning. 
