Evaluate $\int_{0}^{\infty} \frac{\sinh(ax)}{e^{bx} - 1}\ dx$ Is there a simple method to evaluate the following integral
$$
\int_{0}^{\infty} \frac{\sinh(ax)}{e^{bx} - 1}\ dx
$$
 A: Denote
$$
I(a,b)=\int\limits_{0}^\infty f_{a,b}(x)dx\qquad
f_{a,b}(x)=\frac{\sinh(ax)}{e^{bx}-1}
$$
Obviously, $I(0,b)=0$ and $I(-a,b)=-I(a,b)$. If $b\leq 0$, or $|a|> b>0$ then $f_{a,b}$ monotonically tends to $\infty$. If $|a|=b>0$, then the limit of $f_{a,b}$ at infinity is a non-zero constant. In both cases $I(a,b)=\infty$. Using this facts we see that it is enough to consider case $0<|a|<b$. Then
$$
I(a,b)=\int\limits_{0}^\infty \frac{\sinh(ax)}{e^{bx}-1}=\{t=ax\}=\int\limits_{0}^\infty\frac{\sinh t}{e^{\frac{b}{a}t}-1}\frac{dt}{a}=\frac{1}{a}J\left(\frac{b}{a}\right)
$$
where 
$$
J(p)=\int\limits_{0}^\infty\frac{\sinh t}{e^{pt}-1}dt,\qquad p>1
$$
Let's begin
$$
\begin{align}
J(p)&=\int\limits_{0}^\infty\frac{\sinh t}{e^{pt}-1}dt
=\int\limits_{0}^\infty\frac{e^{-pt}\sinh t}{1-e^{-pt}}dt
=\int\limits_{0}^\infty e^{-pt}\frac{e^{t}-e^{-t}}{2}\sum\limits_{k=0}^\infty (e^{-pt})^kdt\\
&=\frac{1}{2}\sum\limits_{k=0}^\infty\int\limits_{0}^\infty e^{-pt}(e^{t}-e^{-t}) (e^{-pt})^kdt
=\frac{1}{2}\sum\limits_{k=0}^\infty\left(\int\limits_{0}^\infty e^{(-p+1-pk)t}dt-\int\limits_{0}^\infty e^{(-p-1-pk)t}dt\right)\\
&=\frac{1}{2}\sum\limits_{k=0}^\infty\left(\frac{1}{pk+p-1}-\frac{1}{pk+p+1}\right)
=\frac{1}{2}\sum\limits_{k=0}^\infty\frac{2}{(pk+p)^2-1}
=\frac{1}{2}\sum\limits_{k=1}^\infty\frac{2}{p^2 k^2-1}\\
\end{align}
$$
Now we use the following uquality
$$
\sum\limits_{k=1}^\infty\frac{2z}{z^2-n^2}=\pi\cot\pi z-\frac{1}{z}
$$
Its proof you can find in this post. Then
$$
J(p)=\frac{1}{2}\sum\limits_{k=1}^\infty\frac{2}{p^2 k^2-1}=
-\frac{1}{2p}\sum\limits_{k=1}^\infty\frac{2p^{-1}}{p^{-2}-k^2}=
-\frac{1}{2p}\left(\pi\cot \pi p^{-1}-\frac{1}{p^{-1}}\right)
=\frac{1}{2}-\frac{\pi}{2p}\cot\frac{\pi}{p}
$$
The final result is
$$
I(a,b)=
\begin{cases}
\frac{1}{2a}-\frac{\pi}{2b}\cot\frac{\pi a}{b}\quad&\text{ if }\quad b>|a|>0\\
0\quad&\text{ if }\quad a=0\\
\infty\quad&\text{ otherwise }\quad
\end{cases}
$$
A: Write $\sinh$ as the sum of two exponentials. The integral only converges if the damping by the denominator is stronger than the growth of the numerator, i.e. if $b\gt a$. If so, you can divide through by $\mathrm e^{bx}$ to make the numerator decay exponentially and bring the denominator into a form that you can expand for small $\mathrm e^{-bx}$. You can integrate the series termwise, then combine the linearly decaying terms pairwise to get quadratically decaying terms and thus a convergent series. Then use the series representation of $\coth x$ at $x=0$.
A: With $\int_0^\infty \frac{\sinh \alpha x}{\sinh\beta x}dx= \frac{\pi}{2\beta}\tan\frac{\pi\alpha}{2\beta}$
\begin{align}
\int_{0}^{\infty} \frac{\sinh ax }{e^{bx} - 1}\ dx
=& \ \frac12 \int_{0}^{\infty} e^{-ax}+ \frac {\sinh(a-\frac b2)x}{\sinh\frac {bx }2} \ dx\\
=& \ \frac1{2}\bigg(\frac1a + \frac\pi b \tan\frac{(a-\frac b2)\pi}{b}\bigg)
=\frac1{2a}-\frac\pi{2b}\cot \frac{a\pi}b
\end{align}
