An integral with trigonometric and rational function

Evaluate the integral

$\displaystyle \int_0^{\infty} \frac{x^2-4}{x^2+4} \frac{\sin 2x}{x} \, {\rm d}x$

Solution

Firstly , we begin by noticing the following

$\displaystyle \int_{0}^{\infty} \frac{(x^2-4) \sin 2x}{\left ( x^2+4 \right )x} \, {\rm d}x = \frac{1}{2} \int_{-\infty}^{\infty} \frac{(x^2-4) \sin 2x}{\left ( x^2+4 \right )x} \, {\rm d}x \quad (1)$

Let us now consider the function $\displaystyle f(z)=\frac{(z^2-4) e^{2iz}}{z(z^2+4)}$ as well as the contour

Rendered by QuickLaTeX.com

The function has three simple poles of which only $z=2i$ is included within the contour. The residue at $z=2i$ turns out to be $e^{-4}$ . Thus

$\displaystyle \oint \limits_{\gamma} f(z) \, {\rm d}z = 2 \pi i {\rm Res}\left ( f ;z =2i \right ) = 2 \pi i e^{-4}$

The contribution of the large circle as $R \rightarrow +\infty$ is $0$ whereas the contribution of the small circle as $\epsilon \rightarrow 0$ is $\pi i$ . This can be seen by parametrising the small circle ( $\epsilon e^{it} \; , \; t \in [0, \pi]$ ). Hence:

$\begin{aligned} \frac{2 \pi i }{e^4} = \int_{-\infty}^{\infty} f(z) \, {\rm d}z +i \pi &\Leftrightarrow \left ( \frac{2}{e^4} -1 \right ) \pi i = \int_{-\infty}^{\infty} \frac{\left ( z^2-4 \right ) e^{2iz}}{z \left ( z^4+4 \right )} \, {\rm d}z \\ &\Leftrightarrow \left ( \frac{2}{e^4} -1 \right ) \pi = \int_{-\infty}^{\infty} \frac{\left ( x^2-4 \right ) \sin 2x}{x(x^2+4)} \, {\rm d}x \end{aligned}$