Calculate $I(t,n) = \int_{-\infty}^{\infty} \big( \frac{1}{1-jq} \big)^{n} e^{-jqt} dq$ 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!
 A: 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.
A: If we define the Fourier transform of a function $f$ and its inverse as
$$\mathcal F(f)(\omega)=\int_{\mathbb R}f(x)e^{-j\omega x}dx \,\,\,\text{ and }\,\,\, f(x)=\frac 1 {2\pi}\int_{\mathbb R} \mathcal F(f)(\omega)e^{jx\omega}d\omega$$
then you're looking for the Fourier transform of $f_n(x)=f(x)^n$ where $f(x)=\frac{1}{1-jx}$. 
Because the Fourier transform maps products to convolutions (times $\frac 1 {2\pi}$, given the Fourier definition we adopted), you're looking for the $n$-th self-convolution of $\mathcal F(f)$.
Now, with $H$ denoting the Heaviside step function, we have $$f(x)=\frac{1}{1-jx}=\int_{\mathbb R}e^{-\omega}H(\omega)e^{jx\omega}d\omega$$
As a consequence, the Fourier transform of $f$ is $\omega\rightarrow 2\pi e^{-\omega}H(\omega)$.
This gives us $$I(t, 1)=2\pi e^{-t}H(t)$$
and you can verify that the $n$-th self-convolution is given by $$I(t,n)=2\pi\frac{t^{n-1}}{(n-1)!}e^{-t}H(t)$$
