I would like to evaluate the two integrals $$I_\pm(u) = \frac{1}{2\pi}\int_{-\infty}^\infty\frac{ e^{-i z u}}{z \pm i c }dz $$ where $c$ is a positive real constant. I expect these each to be a real exponential times some Heaviside function, but I'm having trouble showing this. Can anyone offer guidance? I figure this is trivial for someone with more contour integration experience.
So far, concentrating on the integral $I_-(u)$, I attempted to use the semicircle contour depicted below:
Evaluating the residue at $z=ic$, this provides
$$I_-(u) = i e^{uc} - \lim_{R\rightarrow \infty} \frac{1}{2\pi}\int_0^\pi d\theta \frac{iRe^{i\theta} \exp(-i R e^{i\theta} u )}{Re^{i\theta} - ic}$$
I suppose this limit depends on the sign of $u$ and therefore introduces a heaviside function? Shooting in the dark here...
I don't understand why people don't realize rectangular contours are more natural than semicircles.. It is mostly obvious that for $u <0$, $\lim_{R\to \infty} \int_{\partial [-R,R] + i[0,R]} \frac{e^{-izu}}{z-ic}dz = \int_{-\infty}^\infty \frac{e^{-izu}}{z-ic}dz$. Apply the residue theorem to the LHS. For $u>0$ the integrand in the LHS has exponential growth so we look instead at $\lim_{R\to \infty} \int_{\partial [-R,R] - i[0,R]} \frac{e^{-izu}}{z-ic}dz = -\int_{-\infty}^\infty \frac{e^{-izu}}{z-ic}dz$. The LHS is $0$ by Cauchy's integral theorem.