$$
J(a,b)=\int_{0}^{\infty }\frac{\sin(b x)}{\sinh(a x)} dx
$$
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.
Best Answer
Note first that the integrand is even and can be extended to the entire real line. Consider
$$f(z) = \frac{e^{i b z}}{\sinh{a z}}$$
$b>0$, in the upper half complex plane. Then
$$\oint_C dz \: \frac{e^{i b z}}{\sinh{a z}}$$
is equal to $i 2 \pi$ times the sum of the residues within $C$, which we'll take to be a semicircle of large radius in the upper half plane indented at the origin to include the pole at $z=0$. The integral about the semicircular arc goes to zero as this radius goes to infinity because of Jordan's Lemma. Then the integral of $f$ along the real line is the sum of the residues within $C$.
$\sinh{a z}$ has poles at $z=i n \pi/a$, $n \ge 0$. The residue term here is a little tricky, but may be shown to be equal to
$$\text{Res}_{z=i n \pi/a} f(z) = \frac{(-1)^n}{a} e^{-n \pi b/a}$$
The integral along the real line is then the sum of all the residues within $C$:
$$\int_{-\infty}^{\infty} dx \: \frac{e^{i b x}}{\sinh{a x}} =\frac{i 2 \pi}{a} \sum_{n=0}^{\infty} (-1)^n e^{-n \pi b/a} = \frac{i 2 \pi}{a} \frac{1}{1+e^{-\pi b/a}}$$
For the case $b<0$, we use a contour in the lower half plane but indented to exclude the pole at $z=0$ (which was already included before); the result is:
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i b x}}{\sinh{a x}} =\frac{i 2 \pi}{a} \sum_{n=1}^{\infty} (-1)^n e^{-n \pi b/a} = -\frac{i 2 \pi}{a} \frac{1}{1+e^{\pi b/a}}$$
Then we take $1/2$ the sum to get the integral over the positive reals:
$$\int_{0}^{\infty} dx \: \frac{\sin{b x}}{\sinh{a x}} = \frac{\pi}{2 a} \tanh{\left(\pi \frac{b}{2 a}\right)}$$