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I would like to integrate $\int_0^{\infty}\frac{\csc(a x) \sin(a x m)}{\cosh(x) \exp(x)}\mathrm{d}x$ where $m$ is an integer.

There appears to be singularities both real $x = \frac{n\pi}{a}$ and imaginary $x = \frac{\pi}{2 I} +I \pi n$.

This seems to suggest contour integration is the way to go.

Now I'm not sure how to proceed from here on.

metamorphy
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onepound
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1 Answers1

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For $m>0$, $\displaystyle\frac{\sin axm}{\sin ax}=\sum_{k=0}^{m-1}\cos(m-1-2k)ax$, so the given integral is a sum of things like $$\int_0^\infty\frac{\cos bx}{1+e^{ax}}\,dx=\frac{1}{2a}\left[f\left(\frac{ib}{a}\right)+f\left(-\frac{ib}{a}\right)\right]\tag{*}\label{mainint}$$ where, for a complex $z$ with $\Re z>-1$, $$f(z)=\int_0^\infty\frac{e^{-zx}}{1+e^x}\,dx\underset{e^{-x}=t}{=}\int_0^1\frac{t^z\,dt}{1+t}=\int_0^1\frac{t^z-t^{z+1}}{1-t^2}\,dt\\\underset{t^2=x}{=}\frac12\int_0^1\frac{x^{(z-1)/2}-x^{z/2}}{1-x}\,dx=\frac12\left[\psi\left(1+\frac{z}{2}\right)-\psi\left(\frac{1+z}{2}\right)\right],$$ with $\psi$ the digamma function (the final equality is shown like it's done here). If we had sine in place of the cosine in $\eqref{mainint}$, the $\psi$'s would reduce because of the reflection formula. With the cosine in place, these don't, as well as in the final result. That's why I don't expect contour integration to yield anything useful.

metamorphy
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  • thank you for working through that. I think you're right - I'll accept your answer if nobody has anything else to add. – onepound Aug 17 '20 at 09:16