Evaluating $ \int_{-1}^{1} \arctan \left(e^{x}\right) d x $

Question

Using the substitution x=ln(t), the integral becomes $$ I=\int_{e^{-1}}^{e} \frac{\operatorname{Arctan}(t)}{t} \cdot d t $$ which has no antiderivative expressed with usual functions.

First, i want to prove that $$\frac{\operatorname{Arctan}(t)}{t} $$ has no antiderivative. I thought about using Darboux theorem, but this function is unfortunately continuous on R.
Can we really evaluate the integral I, even when we don't know an antiderivative of the function? I tried expressing the integral in term of limit of Riemann sums $$ \int_{-1}^{1} \operatorname{arctan}\left(e^{x}\right) \cdot d x=\lim _{n \rightarrow+\infty} \frac{2}{n} \cdot \sum_{k=0}^{n} \operatorname{Arctan}\left(e^{-1+k\left(\frac{2}{n}\right)}\right) $$ but i don't know how to do with k . Usually when we can't find an antiderivative to evaluate an integral, we try to bound the integral by the "mean formula". When applying this to my function i get the following:
As our function f is continuous, $$ \begin{aligned} \exists c \in[-1 ; 1]: \int_{1}^{1} \operatorname{Arctan}\left(e^{x}\right) \cdot d x=\operatorname{Arctan}\left(e^{c}\right) \cdot(1-(-1)) \\&=2 \text { Arctan }\left(e^{c}\right) \\ (-1 \leqslant c \leqslant 1) \Rightarrow\left(e^{-1} \leq e^{c} \leq e\right) & \\ \Rightarrow\left(2 \cdot \operatorname{Arctan}\left(e^{-1}\right)\right.&\leqslant I \leqslant 2 \text { Arctan}\left(e^{1}\right) \end{aligned} $$ . Is this last suggestion sufficient to answer the question?

What methods for evaluating integrals are you allowed to use? Certainly you can evaluate many different types of definite integrals even though you don't have an anti-derivative; the most famous example being ${\int_{-\infty}^{\infty}e^{-x^2}dx}$ — Riemann'sPointyNose, Apr 23 '21 at 20:08
For the problem in the title, use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 — lab bhattacharjee, Apr 23 '21 at 20:17
Note that $\int_{-a}^a f(x),dx = \frac{1}{2}\int_{-a}^a (f(x) + f(-x)),dx$. — Brian Moehring, Apr 23 '21 at 20:19
Hint: there is a trick, $\arctan(e^{-x})=\frac\pi2-\arctan(e^x)$. — , Apr 23 '21 at 20:35
Anti in inverse tangent integral function https://math.stackexchange.com/questions/4105985/how-to-calculate-int-arctanex-dx/4106007#4106007 — Quanto, Apr 23 '21 at 20:36

score 4 · Accepted Answer · answered Apr 23 '21 at 20:33

4

The key point is that for $x\in \mathbb{R}$, $\arctan(e^{-x})=\arctan\left(\frac{1}{e^x}\right)=\frac{\pi}{2}-\arctan{e^x}$.

$$\begin{array}{ccc} \int_{-1}^1\arctan(e^x)dx&=&\int_{-1}^{0}\arctan(e^x)\,dx+\int_{0}^{1}\arctan(e^x)\,dx\\ &=&\int_{1}^{0}-\arctan{e^{-t}}dt+ \int_{0}^{1}\arctan(e^x)\,dx\\ &=&\int_{0}^{1}\arctan(e^x)+\arctan(e^{-x})\,dx\\ &=&\int_{0}^{1}\frac{\pi}{2}\,dx\\ \end{array}$$

Therefore $$\int_{-1}^1\arctan(e^x)dx=\frac{\pi}{2}$$

answered Apr 23 '21 at 20:33

Ayoub

1,456

I have completely forgotten this relation! So, in general, when we are evaluating integrals which have no antiderivative we can try to find a relation of this kind? But is there too many integrals of this kind? I remain a bit skeptical..this trick doesn't apply to the @Riemann'sPointyNose example(the gaussian integral) i guess. I am "collecting" methods on integral evaluation and i liked yours. Thanks a lot! – Yagami Apr 23 '21 at 21:26
1

@MouhamadouMoustapha, when you can't find an antiderivative, you can try other methods like integrating by parts, change of variable, use of special properties of the function you're integrating, etc. There's no guarantee that it will work, but sometimes it does. – Ayoub Apr 23 '21 at 21:35

Evaluating $ \int_{-1}^{1} \arctan \left(e^{x}\right) d x $

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