How to integrate $\int \arctan(e^{-\pi y/b}) \, dy$?

Question

How to integrate $\int \arctan(e^{-\pi y/b}) \, dy$

Answer 1

An idea using complex numbers:

Observe that in the complex numbers and up to a constant (and I'm not sure, now of the top of my head, how the multivalued complex logarithm affect this constant):

$$\arctan x=\frac1{2i}\log\frac{x-i}{x+i}\implies$$

$$\arctan e^{-\pi y/b}=\frac1{2i}\log\frac{e^{-\pi y/b}-i}{e^{-\pi y/b}+i}$$

and for example, with

$$\;z:=e^{-\pi y/b}\pm i\implies \log z=\log|z|+i\arg z=\frac12\log\left(e^{-2\pi y/b}+1\right)\pm i\theta\;,\;\;0<\theta<\frac\pi 2\;$$

if we take the principal value of the argument, so we'd get

$$\arctan e^{-\pi y/b}=\frac1{2i}\left(-2i\theta\right)=-\theta$$

and now just express $\;\theta\;$ as a function of $\;y\;$ (this is an arctangent, of course, but depends also on $\;x\;$