Evaluate the integral:
$$\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, {\rm d}x$$
Solution
We have successively:
\begin{align*}\int_0^{\infty} \frac{\sin^2 \tan x}{x^2}\,{\rm d}x&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin^2 \tan x}{x^2}\,{\rm d}x\\&= \frac{1}{2}\sum\limits_{n=-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan (x+n\pi)}{(x+n\pi)^2}\,{\rm d}x\\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \sin^2 \tan x \left(\sum\limits_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}\right)\,{\rm d}x\\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan x}{\sin^2 x}\,{\rm d}x\\&= \int_{0}^{\pi/2} \frac{\sin^2 \cot x}{\cos^2 x}\,{\rm d}x\\&= \int_0^{\infty} \sin^2 \frac{1}{y}\,{\rm d}y = \frac{\pi}{2}\end{align*}
since $\displaystyle \frac{1}{\sin^2 x} = \sum\limits_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}, \; x \notin \mathbb{Z}$ holds.
The exercise can also be found at mathimatikoi.org
$$\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, {\rm d}x$$
Solution
We have successively:
\begin{align*}\int_0^{\infty} \frac{\sin^2 \tan x}{x^2}\,{\rm d}x&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin^2 \tan x}{x^2}\,{\rm d}x\\&= \frac{1}{2}\sum\limits_{n=-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan (x+n\pi)}{(x+n\pi)^2}\,{\rm d}x\\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \sin^2 \tan x \left(\sum\limits_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}\right)\,{\rm d}x\\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan x}{\sin^2 x}\,{\rm d}x\\&= \int_{0}^{\pi/2} \frac{\sin^2 \cot x}{\cos^2 x}\,{\rm d}x\\&= \int_0^{\infty} \sin^2 \frac{1}{y}\,{\rm d}y = \frac{\pi}{2}\end{align*}
since $\displaystyle \frac{1}{\sin^2 x} = \sum\limits_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}, \; x \notin \mathbb{Z}$ holds.
The exercise can also be found at mathimatikoi.org
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