3

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 ?

GNUSupporter 8964民主女神 地下教會
  • 17,627
Atmos
  • 7,369

2 Answers2

7

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 $[0,1)$.

Shashi
  • 8,738
  • (+1) I didn't delete my answer because it is not exactly the same, but if you think it is too close, I will. – robjohn Jan 30 '18 at 23:18
  • There is no need to! How can I think that while yours has a shortcut?! (+1) for yours too! – Shashi Jan 30 '18 at 23:20
5

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

robjohn
  • 345,667