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[SOLVED] residue calculus


Well-known member
Feb 1, 2012
\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
Jan 31, 2012
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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