Solving this complicated integral using the Residue Theorem The following is an integral I am trying to evaluate
$$I= \int_{-\infty}^\infty f(s) \, ds = \int_{-\infty}^\infty \frac{\frac{1}{(1- \  \ 2 \pi j s  )^{m}}-1}{2\pi j s }\ e^{-2\pi j s \ \theta}\ ds $$
where $\theta$ is non-negative constant and $m$ is an positive integer.
Someone helped me by saying that, I can solve it exactly by closing the integration contour in the lower half complex plane, i.e using the Residue formula. Therefore I tried that below
What I tried
Using the  Residue Theorem "Residue Theorem"  and knowing that the pole is at $z^*=\frac{1}{2\pi j}$ then
$$I = -2 \pi j \ \ \text{Res}_{z^*= \frac{1}{2\pi j}}[f(z)]$$
Next I evaluate the residue "residue of function", then
\begin{align}\text{Res}_{z^*= \frac{1}{2\pi j}}\left[f(z)\right]=&\lim_{z\rightarrow z^*} (z-z^*) \frac{\frac{1}{(1- \  \ 2 \pi j z  )^{m}}-1}{2\pi j z }\ e^{-2\pi j z\theta}\\ \\
&=\lim_{z\rightarrow z^*}\ (z-z^*) \frac{\frac{1}{(1- \frac {z}{z*}  )^{m}}-1}{2\pi j z }\ e^{-2\pi j z \theta}\\
&=\lim_{z\rightarrow z^*} (z-z^*) \frac{\frac{(z^*)^m}{(z^*-{z}  )^{m}}-1}{2\pi j z }\ e^{-2\pi j z\theta}\\
&=\lim_{z\rightarrow z^*}\frac{\frac{- (z^*)^m}{(z^*-{z}  )^{m-1}}-1}{2\pi j z }\ e^{-2\pi j z\theta}\\
\\ &= ????
\end{align}
I don't know if I am doing the right thing, do you think my derivation is correct?
Am I at least on the right track?
Thanks
 A: You are using the formula for the residue in the case of a simple pole, but here we have a pole of order $m$ (which may be $1$, but generally isn't), so the formula to use is
$$\operatorname{Res}_{z^\ast = \frac{1}{2\pi j}} [f(z)] = \lim_{z\to z^\ast} \frac{1}{(m-1)!}\biggl(\frac{d}{dz}\biggr)^{m-1} (z-z^\ast)^mf(z).\tag{1}$$
When we write $f$ in the form
$$f(z) = \frac{g(z)}{(z-z^\ast)^m},$$
a Taylor expansion of $g$ shows that
$$f(z) = \sum_{n=0}^\infty \frac{g^{(n)}(z^\ast)}{n!}\cdot (z-z^\ast)^{n-m},$$
and the residue is the coefficient of $(z-z^\ast)^{-1}$, so the term for $n = m-1$, which is
$$\frac{g^{(m-1)}(z^\ast)}{(m-1)!}.$$
Here, we have
$$f(z) = \frac{\frac{1}{(1-2\pi jz)^m}-1}{2\pi j z}e^{-2\pi jz\theta} = \frac{\frac{1-(1-2\pi jz)^m}{2\pi jz}}{(1-2\pi jz)^m}e^{-2\pi j z\theta} = \frac{(-1)^m\frac{1-(1-2\pi jz)^m}{(2\pi j)^m(2\pi jz)}e^{-2\pi jz\theta}}{\bigl(z-\frac{1}{2\pi j}\bigr)^m},$$
so
$$g(z) = (-1)^m\frac{1-(1-2\pi jz)^m}{(2\pi j)^m(2\pi jz)}e^{-2\pi jz\theta}.$$
Since $(1-2\pi jz)^m$ has a zero of order $m$ at $z^\ast = \frac{1}{2\pi j}$, that part of $g$ doesn't contribute to the residue, and we only need to compute the derivatives of
$$\frac{1}{2\pi jz} e^{-2\pi j z\theta}$$
and multiply with the constant factors, which makes the computation a bit simpler.
