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

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

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$• Thank you! It is very difficult to see the obvious one. – user339807 Jan 11 '17 at 16:26 • what is the meaning of $$\frac{1}{n}$$ in front of the integral? – Dr. Sonnhard Graubner Jan 11 '17 at 16:27 •$\arctan(1)$in (2) is wrong. – xpaul Jan 11 '17 at 16:36 • (1) is wrong too. For example, if$n=1$, you get$\int_0^\infty\arctan(\tan x)dx=\int_0^\infty xdx$which is divergent. – xpaul Jan 11 '17 at 16:39 • @xpaul$\arctan(\tan(x)) \neq x$if$x> \frac{\pi}{2}$. – Carl Schildkraut Jan 11 '17 at 17:08 ## 1 Answer 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$(1)reveals \begin{align} I_n&=\int_{-\infty}^\infty \frac{\arctan(e^{nv})}{\cosh(v)}\,dv\\\\ &=\int_{-\infty}^0 \frac{\arctan(e^{nv})}{\cosh(v)}\,dv+\int_0^\infty \frac{\arctan(e^{nv})}{\cosh(v)}\,dv\\\\ &=\int_0^\infty \frac{\arctan(e^{nv})+\arctan(e^{-nv})}{\cosh(v)}\,dv\\\\ &=\int_0^\infty \frac{\pi/2}{\cosh(v)}\,dv\\\\ &=\frac{\pi}{2}\left.\left(2\arctan(e^v)\right)\right|_{v=0}^{v\to \infty}\\\\ &=\left(\frac{\pi}{2}\right)^2 \end{align} as was to be shown! NOTE: The integral as given by(1)$does not equal the integral$\int_0^\infty \arctan(\tan^n(x))\,dx\$, which actually diverges.

• (+1) @Dr.MV. I just realise that mistake. Long tired day. – user339807 Jan 11 '17 at 18:18
• @adbi Thank you for the up vote. Much appreciative. -Mark – Mark Viola Jan 11 '17 at 21:58