Why can't I just find the residue of the function?

Question

I was solving the contour integral $$\oint \frac{z\sin z}{z^{2}+4} \ dz$$ in the upper half of the complex plane using the residue theorem and I couldn't figure out why I needed to convert it to $$Im\left ( \oint \frac{ze^{iz}}{z^{2}+4} \ dz \right )$$ in order to get the correct result of $\frac{\pi }{e^{2}}$. When I use it on the original function I get $-\pi \sinh 2$. Can anyone explain why it doesnt work with the sine function? Thanks.

Answer 1

When you're doing contour integration, you need to be worried that your function decays sufficiently rapidly that the integrals around all your pieces stay bounded, otherwise you may not be able to do the $R \to \infty$ limit or similar that you need to do to finish the computation. What goes awry here is that $\sin(z)$ doesn't decay anywhere: it blows up exponentially fast as $\mathrm{Im}(z)$ grows in either direction. This means that you have a highly nontrivial problem in trying to understand the asymptotics of the integral along the circular arc.

But for $e^{iz}$, it decays exponentially fast with $\mathrm{Im}(z)$ in the upper half plane, so the integral along the arc goes to zero, and you can then take the imaginary part at the end to get the desired result. You could just as well use $\sin(z)=-\mathrm{Im}(e^{-iz})$ with a contour in the lower half plane, if you wanted.