# Evaluate $\int_{0}^{\infty} \frac{\sinh(ax)}{e^{bx} - 1}\ dx$

Is there a simple method to evaluate the following integral $$\int_{0}^{\infty} \frac{\sinh(ax)}{e^{bx} - 1}\ dx$$

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This is essentially a duplicate of a question about a whole class of such integrals that was answered in great generality; unfortunately I can't find it despite a lot of searching. – joriki Nov 4 '12 at 13:31
are there any assumptions made abut the values of $a$ and $b$? – Valentin Nov 4 '12 at 13:38
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 $$I(a,b)= \begin{cases} \frac{1}{2a}-\frac{\pi}{2b}\cot\frac{\pi a}{b}\quad&\text{ if }\quad b>|a|>0\\ 0\quad&\text{ if }\quad a=0\\ \infty\quad&\text{ otherwise }\quad \end{cases}$$
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 $x=0$.