How to evaluate this integral
$$
\int_{-\infty}^\infty\frac{\sin x}{x(x^2 + 1)}dx
$$
I am having a problem to solve this because of two poles when I solve it by integration first from $-R$ to $R$ and then along a semicircle in the upper half plane.
Answer
According to your choice of the contour: $C$ is the upper half-plane and $\Gamma$ is the semicircular arc of radius $R$ (say).
$$\int_C \frac{\sin z}{z(z^2+1)}\mathrm dz=\text{P.V.} \int_{-\infty}^{+\infty} \frac{\sin x}{x(x^2+1)}\mathrm dx+\lim_{R \to \infty}\int_{\Gamma}\frac{\sin z}{z(z^2+1)}\mathrm dz$$
Now, use the fact that if $f(z)=\frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $\text{Res}(f(z), z_0) = \frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $\sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $\Gamma$, use the ML-inequality and you should be good to go.
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