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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


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

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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}

Using (1) we get that

\displaystyle \int_{0}^{\infty} \frac{(x^2-4) \sin 2x}{x(x^2+4)} \, {\rm d}x = \pi e^{-4} - \frac{\pi}{2}

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