Consider the complex integral (it's a probability density)

$\displaystyle
$
$\displaystyle \int^{\infty}_{-\infty} \frac{e^{-i x a}}{(1+i x
b)^{k_{1}}(1-i x \frac{c}{N})^{N k_{2}}}dx
$ for
$\displaystyle a \geq 0$

I need to find the residue at the pole $\displaystyle x_{0}=i/b$
(of order $\displaystyle k_{1}$).
According to the theory, that residue is:

$\displaystyle Res_{x_{0}}=\frac{1}{(k_{1}-1)!} \frac{d^{k_{1}-1}}{dx^{k_{1}-1}} \, [ f(x) (1+i x b)^{k_{1}}]_{x=x_{0}}$

I found the residue. Now, the integral should equal to

$\displaystyle 2 \pi i Res_{x_{0}}$

Please could you say where I did a mistake? Because the result I got is not a probability density...
Thank you!