I am trying to calculate integrals of the form:
$$ I(t, n) = \int_{-\infty}^{\infty} \Big(\frac{1}{1-jq}\big)^{n} e^{-jqt} dq $$
where $j = \sqrt{-1}$. In the case when $n=1$, I have:
$$ I(t, 1) = \int_{-\infty}^{\infty} \frac{1}{1-jq} e^{-jqt} dq $$
Now, my thought was to view this as a function of $t$, and use the Feynmann trick. Ie:
$$ \frac{d}{dt} I(t, 1) = \int_{-\infty}^{\infty} \frac{\partial}{\partial t} \big(\frac{1}{1-jq} e^{-jqt} \big) dq $$
$$ \frac{d}{dt} I(t, 1) = \int_{-\infty}^{\infty} \frac{-jq}{1-jq} e^{-jqt} dq $$
This looks promising, but I can't get it to go anywhere. An even more promising avenue seems to be to express the exponential as a power series around zero. This gives:
$$ I(t, 1) = \int_{-\infty}^{\infty} \frac{1}{1-jq} \sum_{k=0}^{\infty} \frac{(-jqx)^{k}}{k!} dq $$
Since the complex exponential is entire, we can interchange the sum and integral, giving:
$$ I(t, 1) = \sum_{k=0}^{\infty} \frac{x^{k}}{k!} \int_{-\infty}^{\infty} \frac{(-jq)^{k}}{1-jq} dq $$
However, I get stuck here too because I can't find an antiderivative for the integrand.
Any ideas? Am I missing some fundamental theoretical concept or trick or technique that makes all of this difficulty disappear?
Unfortunately, I am really not that good at integration, but I want to get better!
Best Answer
For simplicity we rename $j\to i, q \to -z $ to rewrite the integral as $$\begin{align} \int_{-\infty}^\infty\frac{e^{itz}dz}{(1+iz)^n}&=\int_\Gamma\frac{e^{itz}dz}{(1+iz)^n}=2\pi i\;\underset{z=i}{\text {Res}}\,\frac{e^{itz}}{(1+iz)^n}\\ &=\frac {2\pi}{ i^{n-1}}\left.\frac1 {(n-1)!}\frac {d^{n-1}}{dz^{n-1}} e^{itz}\right|_{z=i}=2\pi\frac{ t^{n-1}e^{-t}}{(n-1)!},\end{align} $$ where $\Gamma $ is the usual counterclockwise-oriented contour consisting of the real axis and a large semicircle in the upper complex half-plane. $t$ is assumed to be positive real number.