Today I ran into a peculiar problem when trying to perform the inverse fourier transform of $\frac{1}{a+jw}$ where a is some number
$$ \mathcal{F^{-1}}(\frac{1}{a+jw}) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} \frac{1}{a+jw} \, e^{jwt} \, \mathrm{d} w $$
Then let's multiple top and bottom by $2\pi i$
$$ \mathcal{F^{-1}}(\frac{1}{a+jw}) = \frac{2\pi i}{2\pi} \frac{1}{2\pi i}\int\limits_{-\infty}^{\infty} \frac{1}{a+jw} \, e^{jwt} \, \mathrm{d} w $$
On the right hand side we have the residue on the contour $[-\infty, \infty]$
Then $$ \mathcal{F^{-1}}(\frac{1}{a+jw}) = i Res(e^{jwt};aj) $$
Finally, we have $$\mathcal{F^{-1}}(\frac{1}{a+jw}) = i e^{-at}$$
But wait a minute, where did our "i" come from?
Can someone verify whether I have properly formulated the inverse transform as a residue? If so, can someone spot the mistake?
You have to take the $j$ out of $a+jw$, i.e.
$\frac{1}{a+jw} = \frac{-j}{-ja+w}$.
Then you can also take $-j$ out of the integral and this factor cancels with $j$. The j is the imaginary unit. Recall: $\oint \frac{f(w)}{w-a} dw = 2 \pi i f(a)$.
Of course you have a residue!