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[SOLVED] calculus of residue two


Well-known member
Feb 1, 2012
\int_0^{\infty}\frac{\cos(mx)}{(x^2 + a^2)^2}dx = \frac{\pi}{4a^3}e^{-am} (1 + am)
The integral is even so
\frac{1}{2}\text{Re}\int_{-\infty}^{\infty}\frac{e^{imz}}{(z + ia)^2(z - ia)^2}dz.
Since the singularity is of order two, I believe I need to use
\int\frac{f'}{f} = 2\pi\sum(\text{numer of zeros} - \text{number of poles})
but I am not sure on how.

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
If $f(z)$ has a pole of order $n$ at $z=z_{0}$, then [tex]\text{Res}[f(z), z_{0}] = \lim_{z \to z_{0}} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} (z-z_{0})^{n} f(z) [/tex].

[tex] \text{Res} \Big[ \frac{e^{imz}}{(z^{2}+a^{2})^{2}}, ia \Big] = \lim_{z\to ia} \frac{d}{dz} (z-ia)^{2} \frac{e^{imz}}{(z^{2}+a^{2})^{2}} [/tex]

[tex] = \lim_{z \to ia} \frac{d}{dz} \frac{e^{imz}}{(z+ia)^{2}} = \lim_{z \to ia} \frac{ime^{imz}(z+ia)^{2} - 2e^{imz}(z+ia)}{(z+ia)^{4}}[/tex]

[tex] = \frac{ime^{-am}(-4a^{2})-2e^{-am}(2ia)}{16a^{4}} = -\frac{i e^{-am}(1+am)}{4a^{3}} [/tex]

So for the same reason as the other problem,

[tex] \int_{-\infty}^{\infty} \frac{e^{imx}}{(x^{2}+a^{2})^{2}} \ dx = 2 \pi i \Big(-\frac{i e^{-am}(1+am)}{4a^{3}} \Big) = \frac{\pi}{2a^{3}} e^{-am} (1+am)[/tex]

Sometimes calculating the residue from that definition is insanely tedious and actually finding the Laurent series expansion is easier.