Arctan integral $ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}$

Question

Is there a closed form for the integral $$ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}$$ for $\forall k \ge 1 $?

Well, I was able to get the closed form for the case where $|k|\le1$, and it is of the form $$ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}=\frac{1}{2k}\left(\textrm{Li}_{2}(k)-\textrm{Li}_{2}(-k)-2\log(k)\tanh^{-1}(k)\right)$$ I would highly apperciate if any one could come up with a technique to solve the above mentioned problem

Answer 1

Integrate by parts $$ \int_0^\infty \frac{\arctan x}{x^2 + k^2} dx = \frac 1k \left. \arctan x \arctan \frac xk \right|_0^\infty - \frac 1k \int_0^\infty \frac{\arctan \frac xk}{x^2 + 1} dx, $$ and then substitute $x = kt $: $$ \frac 1k \int_0^\infty \frac{\arctan \frac xk}{x^2 + 1} dx = \frac{1}{k^2}\int_0^\infty \frac{\arctan t}{t^2 + \frac{1}{k^2}} dt. $$ So now you can use your expression.

In case $k = 1$ we get a recurrent integral.

Answer 2

For $k>0$, \begin{align}J(k)&=\int_0^\infty \frac{\arctan x}{x^2+k^2}dx\\ &\overset{y=\frac{x}{k}}=\frac{1}{k}\int_0^\infty \frac{\arctan(ky)}{1+y^2}dy\\ K(k)&=\int_0^\infty \frac{\arctan(ky)}{1+y^2}dy\\ K^\prime(k)&=\int_0^\infty\frac{y}{(1+y^2)(1+k^2y^2)}dy\\ &=\frac{\ln k}{k^2-1}\\ \int_1^k K^\prime(t)dt&=K(k)-K(1)\\ &=K(k)-\frac{\pi^2}{8}\\ &=\int_1^k\frac{\ln t}{t^2-1}dt\\ J(k)&=\frac{\pi^2}{8k}+\frac{1}{k}\int_1^k\frac{\ln t}{t^2-1}dt \end{align} And the integral can be expressed in terms of log and dilog

Answer 3

Note $ \int_{0}^{\infty}\frac{\tan^{-1}x}{x^{2}+k^{2}} \overset{x=k\sqrt t}= \frac1{2k} J(k) $, where $$J(k)=\int_{0}^{\infty}\frac{\tan^{-1}(k\sqrt t)}{(t+1)\sqrt t}dt,\>\>\> J’(k)= \int_{0}^{\infty}\frac{1}{(t+1)(1+k^2 t)}dt=\frac{\ln k^2}{k^2-1} $$

Then, for $k\ge 1$ \begin{align} \int_{0}^{\infty}\frac{\tan^{-1}x}{x^{2}+k^{2}} =&\frac1{2k}\left( J(\infty) - \int_k^\infty J’(s) ds\right)\\ =& \frac1{2k}\left( \frac{\pi^2}2- 2\int^\infty_k \frac{\ln s}{s^2-1} ds\right) \\ \overset{s=\frac kt }=&\frac1{2k}\left(\frac{\pi^2}2-2\int_0^1 \frac{k\ln t}{k^2-t^2}dt + 2\int_0^1 \frac{k\ln k}{k^2-t^2}dt \right)\\ =& \frac1{2k}\left( \frac{\pi^2}2-\text{Li}_2(\frac1k)+\text{Li}_2(-\frac1k) -2\ln k\coth^{-1}k \right) \end{align}