Integral with trigonometric

Let $a>0$. Evaluate the integral:

$$\int_0^\infty \frac{\sin^2 x}{x^2 \left(x^2+a^2 \right)}\, {\rm d}x$$

Solution

First we note that:

$$\begin{aligned}
\int_{0}^{\infty}\frac{\sin^2 x}{x^2 \left (x^2+a^2 \right )}\, {\rm d}x &=\frac{1}{a^2} \int_{0}^{\infty}\sin^2 x \left ( \frac{1}{x^2}- \frac{1}{a^2+x^2} \right) \, {\rm d}x \\
&= \frac{1}{a^2} \left[\int_{0}^{\infty}\frac{\sin^2 x}{x^2}\, {\rm d}x - \int_{0}^{\infty}\frac{\sin^2 x}{a^2+x^2}\, {\rm d}x\right] \\
&=\frac{1}{a^2} \left[\frac{\pi}{2} - \frac{1}{2}\int_{0}^{\infty}\frac{1-\cos 2x}{a^2+x^2}\, {\rm d}x\right] \\
&= \frac{1}{a^2} \left[ \frac{\pi}{2} - \frac{1}{2}\int_{0}^{\infty}\frac{{\rm d}x}{x^2+a^2} + \frac{1}{2}\int_{0}^{\infty}\frac{\cos 2x}{x^2+a^2}\, {\rm d}x \right] \\
&= \frac{1}{a^2}\left [ \frac{\pi}{2} - \frac{\pi}{4a} + \frac{1}{2}\int_{0}^{\infty}\frac{\cos 2x}{x^2+a^2}\, {\rm d}x \right] \\
\end{aligned}$$

So we have to evaluate the last integral. Noting that the integrand is even we have that:

\begin{equation} \int_{0}^{\infty}\frac{\cos 2x}{x^2+a^2}\, {\rm d}x = \frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos 2x}{x^2+a^2}\, {\rm d}x  \end{equation}

Now consider the function $\displaystyle f(z)= \frac{e^{2iz}}{z^2+a^2}$ that is meromorphic in the UHP. (upper half plane) In the UHP we have that:

$$\begin{aligned}
\left | f(z) \right | &=\left | \frac{e^{2i(x+iy)}}{z^2+a^2} \right | \\
 &= \left |\frac{e^{-2y}\left ( \cos 2x +i \sin 2x \right )}{z^2+a^2} \right |\\
 &= \frac{e^{-2y}}{\left | z^2+a^2 \right |} \xrightarrow{z\rightarrow +\infty } 0
\end{aligned}$$

Also it has a simple pole at $z=ia$ (always in the UHP) and the residue there is:

$$\mathfrak{Res}\left ( f;z=ia \right )=\frac{e^{-2a}}{2ia}$$

Hence if we consider the contour integral where as a contour $\mathcal{C}$ we pick a semicircle of radius $R$ , then we have that:

$$\oint \limits_{\mathcal{C}} f(z)\, {\rm d}z = 2\pi i \mathfrak{Res}\left ( f(z); z=ia \right )= 2\pi i \frac{e^{-2a}}{2ia}= \frac{\pi e^{-2a}}{a}$$

So,

$$\int_{-\infty}^{\infty}\frac{\cos 2x}{x^2+a^2}\, {\rm d}x = \frac{\pi e^{-2a}}{a}\overset{(1)}{\Leftrightarrow} \int_{0}^{\infty}\frac{\cos 2x}{x^2+a^2}\, {\rm d}x = \frac{\pi e^{-2a}}{2a}$$

Therefore:

$$\int_{0}^{\infty}\frac{\sin^2 x}{x^2 \left ( x^2+a^2 \right )}\, {\rm d}x = \frac{1}{a^2} \left[ \frac{\pi}{2} - \frac{\pi}{4a} + \frac{\pi e^{-2a}}{4a} \right] $$