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