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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
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Please consider rewriting your post. – Did Nov 4 '12 at 13:44
I'm really sorry for not clarifying enough about the problem. Indeed I remember this was asked in other non-English website several months ago. I think the answer was eventually like Norbert's result but with no proof. Sorry again for confusion. – Tariq Nov 4 '12 at 21:51
up vote 4 down vote accepted

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 $$ I(a,b)= \begin{cases} \frac{1}{2a}-\frac{\pi}{2b}\cot\frac{\pi a}{b}\quad&\text{ if }\quad b>|a|>0\\ 0\quad&\text{ if }\quad a=0\\ \infty\quad&\text{ otherwise }\quad \end{cases} $$

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Thanks a lot for this comprehensive analysis. It seems not as easy as I expected, and I'm not sure if there are other ways to do so. I would like to keep this question open for some time before closing with your wonderful answer. – Tariq Nov 4 '12 at 22:07
@Tariq, thank you! Let's may be there is a shorter proof. – Norbert Nov 4 '12 at 22:10

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 $x=0$.

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