# [SOLVED]calculus of residue two

#### dwsmith

##### Well-known member
$\int_0^{\infty}\frac{\cos(mx)}{(x^2 + a^2)^2}dx = \frac{\pi}{4a^3}e^{-am} (1 + am)$
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
If $f(z)$ has a pole of order $n$ at $z=z_{0}$, then $$\text{Res}[f(z), z_{0}] = \lim_{z \to z_{0}} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} (z-z_{0})^{n} f(z)$$.

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

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

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

So for the same reason as the other problem,

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

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