# [SOLVED]residue calculus

#### dwsmith

##### Well-known member
$\int_0^{\infty}\frac{x\sin(mx)}{x^2 + a^2}dx = \frac{\pi}{2}e^{-am}$
The inetgral is even so
$\frac{1}{2}\int_{-\infty}^{\infty}\frac{x\sin(mx)}{x^2 + a^2}dx.$
We can also write $$x^2 + a^2$$ as $$(x + ai)(x - ai)$$. Should I also write $$\sin(mx) = \frac{1}{2i}(z^m - 1/z^m)$$? I tried this but it didn't go anywhere.
$\frac{1}{2}\int_{-\infty}^{\infty}\frac{z\sin(mx)}{(z + ai)(z - ai)}dz.$
How do I finish this problem?

#### Random Variable

##### Well-known member
MHB Math Helper
Due to the erratic behavior of $\sin z$ on the complex plane, you want to evaluate either

$\displaystyle \frac{1}{4i} \int_{-\infty}^{\infty} \frac{x(e^{imx}-e^{-imx})}{x^{2}+a^{2}} \ dx$ or $\displaystyle\frac{1}{2} \text{Im} \int_{-\infty}^{\infty} \frac{x e^{imx}}{x^{2}+a^{2}} \ dx$.

The latter is preferable since the former would require splitting the integral into two integrals and using two different contours.

Let $\displaystyle f(z) = \frac{z e^{imz}}{z^{2}+a^{2}}$ and integrate around a contour that consists of the real axis and an infinitely large semicircle in the upper-half plane.

According to Jordan's lemma, the integral evaluates to zero along the semicircle.

Then $\displaystyle \int_{-\infty}^{\infty} \frac{x e^{imx}}{x^{2}+a^{2}} \ dx = 2 \pi i \ \text{Res}[f(z),ia]$

$\displaystyle = 2 \pi i \lim_{z \to ia} \frac{z e^{imz}}{z+ia} = 2 \pi i \frac{ia e^{-am}}{2ia} = i \pi e^{-am}$

And the result follows.

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