# Integral $\int_{-\infty}^{\infty} \arctan(e^x) \arctan(e^{-x})dx=\frac{7}{4}\zeta(3)$

How to prove that $$\int_{-\infty}^{\infty} \arctan(e^x) \arctan(e^{-x})\rm dx=\frac{7}{4}\zeta(3)?$$

This integral appeared while attempting to solve an unanswered question, namely:

Closed form of :$\int_{-\infty}^{\infty}\arctan\left(e^{-x^2 \text{erf}(x)}\right)\,\arctan\left(e^{x^2\text{erf(x)}}\right)\,dx$

I happened to find that $$\int_{-\infty}^{\infty} \arctan(e^x) ~\arctan(e^{-x})~dx=\frac{7}{4}\zeta(3).$$

How can you work out the answer on the right by hand?

I guess that there could be multiple approaches to solve this one.

• What's the question? Jun 17, 2019 at 4:10
• @Lord Shark the Unknown : I suspect, to derive the expression for the integral's value by hand. Jun 17, 2019 at 4:34
• @Loord Shark the Unknown, I will submit a proof shortly. Jun 17, 2019 at 5:00
• Please do, this one is really fascinating Jun 17, 2019 at 5:06
• I'm voting to close this question as off-topic because this is not a question. Jun 17, 2019 at 6:51

$$\int_{-\infty}^\infty \arctan(e^x)\arctan(e^{-x})dx\overset{e^{x}=t}=\int_0^\infty \frac{\arctan t \arctan(1/t)}{t}dt$$ $$\overset{t=\tan x}=\int_0^\frac{\pi}{2} \frac{x(\pi/2-x)}{\sin x\cos x}dx \overset{2x=t}=\frac14 \int_0^{\pi}\frac{t(\pi-t)}{\sin t}dt\overset{t=\pi x}=\frac{\pi^3}4 \int_0^1 \frac{x-x^2}{\sin(\pi x)}dx$$ Combinging with this post gives us the desired result: $$\boxed{\int_{-\infty}^\infty \arctan(e^x)\arctan(e^{-x})dx=\frac{7\zeta(3)}{4}}$$

• Very good: short and concise. ;) (+1) Jun 17, 2019 at 9:56
• Thanks :D Also your answer from there keeps it that way till the end! Jun 17, 2019 at 10:16
• Thanks for your kind remark! ;) Jun 17, 2019 at 11:10

Let $$I= \int_{-\infty}^{\infty} \arctan(e^x) ~ \arctan(e^{-x}) ~ dx=\int_{0}^{\infty} \frac{\arctan(t) ~ \arctan(1/t)}{t} dt.$$ Let $$t= \tan \theta$$, them $$I=\int_{0}^{\pi/2}2 \theta(\pi/2-\theta)~ \mbox{cosec}2\theta~d\theta=\pi \int_{0}^{\pi/2} \theta ~\mbox{cosec} 2 \theta ~d\theta- \int_{0}^{\pi/2} 2 \theta^2 \mbox{cosec}2\theta ~d\theta$$. Integration by parts gives $$I=\frac{\pi}{2} \left (\left .\theta ~ \ln \tan \theta ~\right|_{0}^{\pi/2}\right)-\frac{\pi}{2} \int_{0}^{\pi/2} \ln \tan \theta ~ d\theta-\left ( \left . \theta^2 ~ \ln \tan \theta ~\right|_{0}^{\pi/2}\right)+\int_{0}^{\pi/2} 2 \theta ~ \ln \tan \theta d\theta.$$ In parenthesis the lower limit $$\lim_{\theta \rightarrow 0}$$ vanishes and at the upper limit these terms in the parentheses diverge and cancell each other. The second terms in the above vanishes because $$\int_{0}^{\pi/2} \ln \sin x dx= \int_{0}^{\pi/2} \ln \cos x dx$$. Then we get $$I=2\int_{0}^{\pi/2} \theta ~\ln \tan \theta ~d\theta= \int_{0}^{\pi/2} \left( -2\theta \ln i -4 \sum_{0}^{\infty} \theta \frac{e^{-2i(2k+1)\theta}}{2k+1}\right) d\theta$$ $$\Rightarrow I =-\frac{i\pi^3}{8}- \sum_{k=0}^{\infty}\left ( 4 \theta \frac{\exp{[-2i(2k+1)\theta]}}{-(2i)(2k+1)^2} \right )_{0}^{\pi/2} -\sum_{k=0}^{\infty} \left( \frac{\exp{[-2i(2k+1)\theta]}}{(2k+1)^3} \right)_{0}^{\pi/2}.$$ $$\Rightarrow I= -\frac{i\pi^3}{8} + i \pi \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}+ 2 \sum_{k=0}^{\infty} \frac{1}{(2k+1)^3}.=-\frac{i\pi^3}{8}+i\pi \frac{\pi^2}{8}+\frac{7}{4} \zeta(3).$$

Here is another slight variation on a theme.

Let $$I = \int_{-\infty}^\infty \arctan (e^x) \arctan (e^{-x}) \, dx.$$ Setting $$u = e^x$$ gives \begin{align} I &= \int_0^\infty \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u}\\ &= \int_0^1 \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u} + \int_1^\infty \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u}\\ &= 2 \int_0^1 \arctan x \arctan \left (\frac{1}{x} \right ) \frac{dx}{x}, \end{align} where in the second of the integrals a substitution of $$x \mapsto 1/x$$ has been enforced.

As $$\arctan \left (\frac{1}{x} \right ) = \frac{\pi}{2} - \arctan x, \qquad x > 0,$$ we can write the above integral as $$I = \pi \int_0^1 \frac{\arctan x}{x} \, dx - 2 \int_0^1 \frac{\arctan^2 x}{x} \, dx = \pi I_1 - 2 I_2.$$

For the first integral $$I_1$$, we have \begin{align} I_1 &= \int_0^1 \frac{\arctan x}{x} \, dx\\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1} \int_0^1 x^{2n} \, dx\\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^2}\\ &= \mathbf{G}, \end{align} where $$\mathbf{G}$$ denotes Catalan's constant.

For the second integral $$I_2$$, integrating by parts gives $$I_2 = -2 \int_0^1 \frac{\ln x \arctan x}{1 + x^2} \, dx.$$ Enforcing a substituition of $$x \mapsto \arctan x$$ leads to $$I_2 = -2 \int_0^{\frac{\pi}{4}} x \ln (\tan x) \, dx.$$ The integral appearing above has been evalauted elsewhere on this site (see, for example, here). The result is: $$\int_0^{\frac{\pi}{4}} x \ln (\tan x) \, dx = \frac{7}{16} \zeta (3) - \frac{\pi \mathbf{G}}{4}.$$ Thus $$I_2 = \frac{\pi \mathbf{G}}{2} - \frac{7}{8} \zeta (3),$$ giving $$I = \pi \mathbf{G} - 2 \left (\frac{\pi \mathbf{G}}{2} - \frac{7}{8} \zeta (3) \right ),$$ or $$\int_{-\infty}^\infty \arctan (e^x) \arctan (e^{-x}) \, dx = \frac{7}{4} \zeta (3),$$ as required.