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.

  1. 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.
  2. 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?
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    $\begingroup$ For the problem in the title, use math.stackexchange.com/questions/439851/… $\endgroup$ Apr 23, 2021 at 20:17
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    $\begingroup$ Note that $\int_{-a}^a f(x)\,dx = \frac{1}{2}\int_{-a}^a (f(x) + f(-x))\,dx$. $\endgroup$ Apr 23, 2021 at 20:19
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    $\begingroup$ Hint: there is a trick, $\arctan(e^{-x})=\frac\pi2-\arctan(e^x)$. $\endgroup$
    – user65203
    Apr 23, 2021 at 20:35
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    $\begingroup$ You mean no elementary antiderivative. $\endgroup$
    – J.G.
    Apr 23, 2021 at 20:36
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    $\begingroup$ Anti in inverse tangent integral function math.stackexchange.com/questions/4105985/… $\endgroup$
    – Quanto
    Apr 23, 2021 at 20:36

1 Answer 1


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

  • $\begingroup$ 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! $\endgroup$
    – Yagami
    Apr 23, 2021 at 21:26
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    $\begingroup$ @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. $\endgroup$
    – Ayoub
    Apr 23, 2021 at 21:35

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