A toughie trigonometric integral

Prove that

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

Solution

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  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*}

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