Jason Yuhas

2021-12-31

Find the value of ${\int}_{-\pi}^{\pi}(4\mathrm{arctan}\left({e}^{x}\right)-\pi )dx$

Ive

Ive

Mary Goodson

Beginner2022-01-01Added 37 answers

Let $f\left(x\right)=4\mathrm{arctan}\left({e}^{x}\right)-\pi$ . Then since $\mathrm{arctan}u+\mathrm{arctan}\frac{1}{u}=\frac{\pi}{2}$ (∗):

$-f(-x)=-4{\mathrm{arctan}e}^{-x}+\pi =-4(\frac{\pi}{2}-{\mathrm{arctan}e}^{x})+\pi =4{\mathrm{arctan}e}^{x}-\pi =f\left(x\right)$

(∗) can be proved geometrically:$\mathrm{arctan}u$ is the angle with opposite side u and adjacent side 1, and $\mathrm{arctan}\frac{1}{u}$ is the other acute angle in the right triangle. Their sum must thus be $\frac{\pi}{2}$ . Alternatively, differentiate the left-hand side and show it is constant, then choose a convenient value for u since any u works, say u=1.

(∗) can be proved geometrically:

autormtak0w

Beginner2022-01-02Added 31 answers

Write the integral as

$I={\int}_{-\pi}^{0}4\mathrm{arctan}\left({e}^{x}\right)-\pi dx+{\int}_{0}^{\pi}4\mathrm{arctan}\left({e}^{x}\right)-\pi dx$

Upon making the substitution u=−x, we find that

$I=-{\int}_{\pi}^{0}4\mathrm{arctan}\left({e}^{-u}\right)-\pi du+{\int}_{0}^{\pi}4\mathrm{arctan}\left({e}^{x}\right)-\pi dx$

$={\int}_{0}^{\pi}4\mathrm{arctan}\left({e}^{-x}\right)-\pi dx+{\int}_{0}^{\pi}4\mathrm{arctan}\left({e}^{x}\right)-\pi dx$

$={\int}_{0}^{\pi}4(\mathrm{arctan}\left({e}^{x}\right)+\mathrm{arctan}\left({e}^{-x}\right))-2\pi dx$

Note that

$\mathrm{arctan}(u+\mathrm{arctan}\left(\frac{1}{u}\right))=\{\begin{array}{cc}\frac{\pi}{2}& \text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}u0\\ -\frac{\pi}{2}& \text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}u0\end{array}$

In this case,$x\in [0,\pi ]$ , so $u={e}^{x}>0$ and $\mathrm{arctan}\left(u\right)+\mathrm{arctan}\left(\frac{1}{u}\right)=\frac{\pi}{2}$ , meaning that the integrand is the zero function. Hence, I=0.

Upon making the substitution u=−x, we find that

Note that

In this case,

karton

Expert2022-01-08Added 556 answers

The answer expands on my comment, which remarked that the integrand of the given integral is twice the Gudermannian function, gd, which appears most famously in the equation governing the Mercator projection in cartography.

NSK

Differentiating the integrand gives

In particular, this derivative is even. Since evaluating the integrand at x=0 gives

for any a: In particular, the occurrence of

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