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The Integral Calculator couldn't help me with the following integral:

$$\int_0^\infty x\frac{\cosh(bx)}{\sinh(x)}dx.$$

From some mathematical physics considerations, I get that the answer should be of the form: $\dfrac{const}{1-cos(b)}$. Is it true? Could you, please, help me with intermediate steps?

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    $\begingroup$ It should go to $\infty$ as $b\to1$, so it's certainly not $C/(1-\cos b)$. $\endgroup$ – Lord Shark the Unknown Jul 16 '18 at 5:54
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    $\begingroup$ $b\in(-1,1)$ atleast for the integral to converge. $\endgroup$ – Sonal_sqrt Jul 16 '18 at 5:59
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Hint: One way is using $\displaystyle\int_0^\infty\frac{\sin ax}{\sinh x}dx=\frac{\pi}{2}\tanh(\frac{\pi a}{2})$, then differentiating respect to $a$ gives an answer with $a=ib$.

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    $\begingroup$ Also the answer is $\dfrac{\pi^2}{4}\dfrac{1}{1-\cos\pi b}$ $\endgroup$ – Nosrati Jul 16 '18 at 7:29

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