$ \int_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \tan^{-1}(e^{\tan(x)}) dx $ any hints?

Question

I am struggling with following integral.

$$ \int_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \tan^{-1}(e^{\tan(x)}) dx $$

A small hint is enough.

Try the substitution $u=-x$ and use the fact that $$\arctan(x)+\arctan(1/x)=\pi/2$$ for any $x>0$. You'll get $\pi^2/8$ as an answer. — Matthew H., Jun 26 '21 at 20:02
Fun fact: You can replace $\tan(x)$ with any odd function $f$ in the exponent of $e$ and you'll always get $\pi^2/8$ as an answer. — Matthew H., Jun 26 '21 at 20:08
@MatthewPilling We can also replace $e$ with any $a>0$, isn't? — azif00, Jun 26 '21 at 20:11

score 4 · Accepted Answer · answered Jun 26 '21 at 20:12

Here is my approach by using the well-known property : $\arctan(f(x)) + \arctan\left(\frac{1}{f(x)}\right) = \frac{\pi}{2}$. One has: $$\int_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \tan^{-1}(e^{\tan(x)}) dx = \int_{- \frac{\pi}{4}}^{0} \tan^{-1}(e^{\tan(x)}) dx+ \int_{0}^{\frac{\pi}{4}} \tan^{-1}(e^{\tan(x)}) dx$$ $$ \overset{t = - x} =\int_{0}^{\frac{\pi}{4}} \tan^{-1}(e^{-\tan(t)}) dt + \int_{0}^{\frac{\pi}{4}} \tan^{-1}(e^{\tan(x)}) dx$$ $$ $$ $$=\int_{0}^{\frac{\pi}{4}} \tan^{-1}(e^{-\tan(x)})+ \tan^{-1}(e^{\tan(x)}) dx= \frac{\pi}{2}\int_{0}^{\frac{\pi}{4}} dx = \frac{\pi^2}{8} $$

$ \int_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \tan^{-1}(e^{\tan(x)}) dx $ any hints?

1 Answers1