Question

How do you take the integral from 1 to infinity of (arctan n) / (1 + n^2) ?

Answer 1

u-substitution:\[\int{\tan^{-1}n\over1+n^2}dn\]\[u=\tan^{-1}n\to du=\frac1{1+n^2}dn\]\[\int udu=\frac12u^2=\frac12(\tan^{-1}n)^2|_{1}^{\infty}\]break this up into a limit integral pair:\[=\lim_{t \rightarrow \infty}\frac12(\tan^{-1}t)^2-\frac12(\tan^{-1}1)^2\]\[\frac12(\frac\pi2)^2-\frac12(\frac{\sqrt2}2)^2=\frac{\pi^2-2}8\]

Answer 2

Oh, I see. How stupid of me.
Thank you so much!

Answer 3

welcome :D