Prove that ${1\over n}\int_{-\infty}^{\infty}{\arctan(e^v)\over \cosh(v/n)}\mathrm dv={\left(\pi\over 2\right)^2}$

Question

$$\int_{0}^{\pi\over 2}\arctan(\tan^n(x))\mathrm dx={\pi^2\over 8}\tag1$$

Is this integral $(1)$ a trivial problem? A easy problem to solve but I can't see it.

$$\int_{0}^{\pi\over 2}\arctan(1)\mathrm dx={\pi^2\over 8}\tag2$$

Why $(1)$ acts like integral $(2)$ for any real values of n?$

An attempt: by enforcing $u=\tan^n(x)$ then $du=n(\tan{x})^{n-1}\sec^2{x}dx$

Recall: $1+\tan^2{x}=\sec^2{x}$

$${1\over n}\int_{0}^{\infty}{\arctan{u}\over u}\cdot{\mathrm dx\over u^{1/n}+u^{1/n}}\tag3$$

Enforcing another substitution: $u=e^v$ then $du=e^vdv$

Recall: $e^x+e^{-x}=2\cosh{x}$

$${1\over 2n}\int_{-\infty}^{\infty}{\arctan(e^v)\over \cosh(v/n)}\mathrm dv\tag4$$

How can we prove $(5)?$

$${1\over n}\int_{-\infty}^{\infty}{\arctan(e^v)\over \cosh(v/n)}\mathrm dv={\left(\pi\over 2\right)^2}\tag5$$

$n\ne 0$

Answer 1

Let $I_n$ be the integral given by

$$I_n=\frac1n\int_{-\infty}^\infty \frac{\arctan(e^v)}{\cosh(v/n)}\,dv \tag 1$$

Enforcing the substitution $v\to v/n$ in $(1)$ reveals

$$\begin{align} I_n&=\int_{-\infty}^\infty \frac{\arctan(e^{nv})}{\cosh(v)}\,dv\\\\ &=\int_{-\infty}^0 \frac{\arctan(e^{nv})}{\cosh(v)}\,dv+\int_0^\infty \frac{\arctan(e^{nv})}{\cosh(v)}\,dv\\\\ &=\int_0^\infty \frac{\arctan(e^{nv})+\arctan(e^{-nv})}{\cosh(v)}\,dv\\\\ &=\int_0^\infty \frac{\pi/2}{\cosh(v)}\,dv\\\\ &=\frac{\pi}{2}\left.\left(2\arctan(e^v)\right)\right|_{v=0}^{v\to \infty}\\\\ &=\left(\frac{\pi}{2}\right)^2 \end{align}$$

as was to be shown!

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NOTE: The integral as given by $(1)$ does not equal the integral $\int_0^\infty \arctan(\tan^n(x))\,dx$, which actually diverges.