I have an integral for which the published solution is very simple, yet Mathematica spits out several lines of junk (much longer than the published result, with references to numerical evaluation of complex integrals).
The integral is
$$\int_{-\infty}^\infty\frac{dp}{2\pi h}\frac{e^{ip(x-x')/h}}{E-\frac{p^2}{2m} + i\epsilon}$$
for which the published result is
$$\frac{-im}{h^2k}e^{\pm ik(x-x')}$$
where the sign of the power in the exponential depends on the sign of $(x-x')$ (this also depends on subbing $E=h^2k^2/2m$, which I do below). They say they rely on "contour integration"... I'm not sure how that applies here, or if there's a simpler way
Mathematica gives the following for the $(x-x')>0$ case:
Physicist's note before we start calculations: every $h$ in here should really be an $\hbar$, but I'll stick with the OP's notation as if the physical context of this problem was unknown.
We'll begin by rewriting the integral with $\epsilon=0$ as $\frac{-m}{\pi h}\int_{\Bbb R}\frac{e^{ip(x-x^\prime)/h}}{p^2-(hk)^2}$. We can now continue by the residue theorem. If $x-x^\prime$ has sign $\pm$, the pole at $p=\pm hk$ contributes$$\pm 2\pi i\frac{-m}{\pi h}\lim_{p\to\pm hk}\frac{e^{ip(x-x^\prime)/h}}{p\pm hk}=\frac{-im}{h^2k}e^{\pm ik(x-x^\prime)}.$$