I am asked to evaluate
$\int_{-\infty}^{\infty} \frac{e^{ixz}}{(1+iz)^2}$, where x is just a constant
Since $e^{izx}$ decays in the upper halfplane, the integral is equal to
$\int_{-\infty}^{\infty} \frac{e^{ixz}}{(1+iz)^2} = 2\pi i \mathrm{Res}f(z)$.
But when calculating the poles i run into som trouble.
$f(z)$ obviously has a pole of second order at z = i, but using the following formula to calcualte the residue:
$\lim_{z \rightarrow i} \frac{d}{dz}((z-i)^2 \frac{e^{ixz}}{(1+iz)^2})$
i run into some problems, as the denumerator still gives a singularity for $z=i$ after differntiation.
Any tips as to how i should proceed?
No, it doesn't. Note that$$\frac{z-i}{1+iz}=-i$$and that therefore$$(z-i)^2\frac{e^{ixz}}{(1+iz)^2}=(-i)^2e^{ixz}=-e^{ixz}.$$