Home » Uncategorized » A toughie trigonometric integral

A toughie trigonometric integral

Prove that

    \[\int_{0}^{\infty} \frac{\sin^2 \tan x}{x^2} \, \mathrm{d}x = \frac{\pi}{2}\]

Solution

We state two lemmata first.

Lemma 1: Let x \in \mathbb{R} \setminus \mathbb{Z}. It holds that

    \[\pi \cot \pi x = \frac{1}{x} + \sum_{n=1}^{\infty}\left ( \frac{1}{x+n} - \frac{1}{x-n} \right )= \lim_{N \rightarrow +\infty}\sum_{n=-N}^{N}\frac{1}{x+n}\]

Proof: A standard proof can be found through Fourier Series. One can expand in Fourier series the function f(x) = \cos ax \; , \; a \notin \mathbb{Z}, \;\; |x| \leq \pi. Another way to prove the identity is to begin from the Weierstrass product , that is \sin \pi x = \pi x \prod \limits_{n=1}^{\infty}\left ( 1- \frac{x^2}{n^2} \right ). Taking log on both sides we have that

    \begin{align*} \ln \sin \pi x &= \ln \pi x + \sum_{n=1}^{\infty}\ln \left ( 1- \frac{x^2}{n^2} \right )\\ &= \ln \pi x + \sum_{n=1}^{\infty}\left [ \ln \left ( 1-\frac{x}{n} \right ) - \ln \left ( 1+ \frac{x}{n} \right ) \right ] \end{align*}

Differentiating we have

    \[\frac{\pi \cos \pi x}{\sin \pi x} = \frac{1}{x}+ \sum_{n=1}^{\infty}\left ( \frac{1}{x+n} - \frac{1}{x-n} \right )= \lim_{N \rightarrow +\infty}\sum_{n=-N}^{N}\frac{1}{x+n}\]

Lemma 2: It holds that

    \[\sum_{n=-\infty}^{\infty}\frac{1}{\left ( z+n \right )^2} = \frac{\pi^2}{\sin^2 \pi z}\]

Proof: Just differentiate the above identity.

Apply the above lemmata , we have for the initial integral that

    \begin{align*} \int_0^{\infty} \frac{\sin^2 \tan x}{x^2}\,\mathrm{d} x&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin^2 \tan x}{x^2}\,\mathrm{d} x \\ &= \frac{1}{2}\sum_{n=-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan (x+n\pi)}{(x+n\pi)^2}\,\mathrm{d} x\\ &= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \sin^2 \tan x \left(\sum_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}\right)\,\mathrm{d} x\\ &= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan x}{\sin^2 x}\,\mathrm{d} x\\ &= \int_{0}^{\pi/2} \frac{\sin^2 \cot x}{\cos^2 x}\,\mathrm{d} x\\ &= \int_0^{\infty} \sin^2 \frac{1}{y}\,\mathrm{d} y \\ &= \frac{\pi}{2} \end{align*}

Read more

1 Comment

  1. Another related Fourier series is the following:

        \[\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n+z}= \frac{\pi}{\sin \pi z}\]

    Proof: We are using the reflection formula of the digamma function to extract the result. We have successively:

        \begin{align*} \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n+z} &= \sum_{n=0}^{\infty} \frac{(-1)^n}{n+z} + \sum_{n=0}^{\infty} \frac{(-1)^n}{z-n} - \frac{1}{z}\\  &= \sum_{n=0}^{\infty} \left [ \frac{1}{z+2n} - \frac{1}{z+2n+1} \right ] + \\ & \quad \quad +\sum_{n=0}^{\infty} \left [ \frac{1}{z-2n} - \frac{1}{z-2n-1} \right ] - \frac{1}{z}\\  &= \frac{1}{2} \left [ \psi^{(0)} \left ( \frac{z+1}{2} \right ) - \psi^{(0)} \left ( \frac{z}{2} \right ) \right ] +\\ &\quad \quad +\frac{1}{2}\left [ \psi^{(0)} \left ( 1- \frac{z}{2} \right ) - \psi^{(0)} \left ( 1- \frac{z+1}{2} \right ) \right ] +\frac{1}{z}-\frac{1}{z} \\  &= \frac{\pi}{2}\left [ \cot \frac{\pi z}{2} +\tan \frac{\pi z}{2} \right ]\\  &= \pi \csc \pi z  \end{align*}

    and the result follows.

    Of course,

        \begin{align*}  \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n+z} = \frac{\pi}{\sin \pi z} &\Rightarrow \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n \pi + z } = \frac{1}{\sin z} \\  &\Rightarrow \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{ n\pi + \frac{\pi}{2}-z} = \frac{1}{\cos z} \end{align*}

Leave a comment

Donate to Tolaso Network