I came across the following integral in my work
$$\int_{-\infty}^{\infty} \frac{\frac{1}{(1- \ \ 2 \pi j s \theta)^{m}}-1}{2\pi j s }\ e^{-2\pi j s\sigma^2}\ ds $$
Assuming $\theta,m,\sigma^2$ are non-negative but nocessarly integers, and $j$ is the imaginary unit.
I tried solving numerically (MATLAB) by replacing the infinite limits by the following $$\int_{-L}^{L} \frac{\frac{1}{(1-\ \ 2 \pi j s \theta)^{m}}-1}{2\pi j s }\ e^{-2\pi j s\sigma^2}\ ds $$
It seems to me there is a singularity point at $s=0$ but the integral still exists even if there is singularity point.
Some tricks have to be done in MATLAB and what I did is, split the integral into two parts at the singularity point and taking the limit over that space to be almost zero.
My question is what would be the best choice of the finite limit $L$? in other words how can one approximate the error from using the approximation? I would like to have the best limits and not have $-\infty$ to $+\infty$ because that would take a whole lot of time.
Any thoughts much appreciated.
In the joint page, it is shown that the result, in case $m>0$ and $\frac{\sigma^2}{\theta}>0$ , is: $$\int_{-\infty}^{\infty} \frac{\frac{1}{(1- \ \ 2 \pi j s \theta)^{m}}-1}{2\pi j s }\ e^{-2\pi j s\sigma^2}\ ds = \frac{\Gamma(m\:,\:\frac{\sigma^2}{\theta})}{\Gamma(m)}$$ The method consists in the change of variables leading to two known Fourier transforms.