I'm trying to evaluate the following integral:
$$\mathcal{J}=\int_{0}^{\infty}\frac{\sin x}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$$
Well there are $3$ poles , one lying on the real line the other on the upper half plane and the other on the lower half plane. The residue at $z=i$ is $\displaystyle -\frac{3}{4e}$ .
I'm integrating on a contour that looks like a semicircle on the upper half plane and has a branch about the origin .
Well I considered the function $\displaystyle f(z)=\frac{e^{iz}+1}{z\left ( z+1 \right )^2}$ which is clearly analytic expect for the poles. Hence if $\gamma$ denotes the contour , then:
$$\oint_{\gamma}f(z)\,{\rm d}z=\oint_{\gamma}\frac{e^{iz}+1}{z\left ( z^2+1 \right )^2}\,{\rm d}z =\oint_{\gamma}\frac{e^{iz}}{z\left ( z^2+1 \right )^2}\,{\rm d}z+\oint_{\gamma}\frac{{\rm d}z}{z\left ( z^2+1 \right )^2}=-2\pi i \frac{3}{4e}+2\pi i = 2\pi i \left ( 1-\frac{3}{4e} \right )=i \left (2\pi - \frac{3\pi}{2e} \right )$$
Hmm... applying the classical method I get that:
$$\int_{-\infty}^{\infty}f(x)\,{\rm d}x =2\pi - \frac{3\pi}{2e}$$
which is almost correct except for that $2$ in front of $\pi$. Where I have gone wrong?
P.S: I used the very obvious that the integrand is even.
Answer
Depict carefully the path of integration: it is a semicircle in the upper half plane with a bulge at $z=0$ and a keyhole around $z=i$. This gives that you have to compute the residues of $f(z)=\frac{e^{iz}}{z(z^2+1)^2}$ at $z=0$ and $z=i$, but to consider only half the residue at $z=0$:
$$\mathcal{J}=\frac{1}{2}\text{Im}\int_{-\infty}^{+\infty}\frac{e^{iz}}{z(z^2+1)^2}\,dz = \frac{1}{2}\text{Im}\left(2\pi i\operatorname{Res}(f(z),z=i)+\pi i\operatorname{Res}(f(z),z=0)\right)$$
so:
$$\mathcal{J} = \frac{1}{2}\text{Im}\left(2\pi i\cdot \frac{-3}{4e}+\pi i\right)=\frac{1}{2}\left(\pi-\frac{3\pi}{2e}\right)=\color{red}{\frac{\pi}{2}\left(1-\frac{3}{2e}\right)}.$$
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