Compute $\int_{0}^{+\infty}\text{arctan}\left(e^{-x}\right)\text{d}x$

Question

I think I've shown that the integral $I$ defined by $$ I=\int_{0}^{+\infty}\text{arctan}\left(e^{-x}\right)\text{d}x $$ exists and I wonder what is its value.

The function $s :x \mapsto \text{arctan}\left(e^{-x}\right)$ is continuous and positive on $\mathbb{R}^{+}$ and $$ \text{arctan}\left(e^{-x}\right)\underset{(+\infty)}{\sim}e^{-x}=o\left(\frac{1}{x^2}\right) $$ The function $\displaystyle x \mapsto \frac{1}{x^2}$ is integrable on $\left[1,+\infty\right[$ so $s$ is integrable on $\left[1,+\infty\right[$ and by continuity on $\left[0,1\right]$, $s$ is integrable on $\left[0,+\infty\right[$.

How can I compute it ?

Answer 1

Set $u=e^{-x}$ so that: \begin{align} I:=\int^\infty_0 \arctan(e^{-x})\,dx=\int^1_0 \frac{\arctan(u)}{u}\,du \end{align} Using taylor series of the $\arctan$ we get: \begin{align} I=\int^1_0 \sum_{k=0}^\infty\frac{(-1)^k}{2k+1} u^{2k}\,du = \sum_{k=0}^\infty \int^1_0 \frac{(-1)^k}{2k+1} u^{2k}\,du = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}=G \end{align} where $G$ is the Catalan's constant. Surely the interchanging of summation and integration must be justified. That can be easily done by using the fact that the series converges uniformly on any compact interval in $[0,1)$.

Answer 2

$$ \begin{align} \int_0^\infty\arctan\left(e^{-x}\right)\,\mathrm{d}x &=\int_0^\infty\sum_{k=0}^\infty(-1)^k\frac{e^{-(2k+1)x}}{2k+1}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\frac1{(2k+1)^2}\\[6pt] &=\mathrm{G} \end{align} $$ As noted by Shashi, $\mathrm{G}$ is Catalan's Constant.