Why are the integrals over $\cos(z)$ and $e^{iz}$ so closely related

Question

Currently I'm studying how to solve improper real integrals using the residue theorem. Usually it goes as follows:

1. Write the improper integral as a limit, e.g. $\int_{-\infty}^\infty f(z)dz=\lim_{R\to\infty}\int_{-R}^Rf(z)dz$.
2. Construct a complex contour containing the real linepiece from above
3. Prove, using the Residue Theorem, that the entire contour integral equals something and prove that the integral over the complex part of the contour goes to zero as $R$ goes to infinity.

Currently I'm working on a homework exercise using the following improper integral: $$\int_{-\infty}^\infty \dfrac{\cos(x)}{(x^2+4)^2(x^2+9)}dx$$ I've tried looking up examples of integrals of such form, i.e. with $\cos(x)$, and they all just say: "We have this integral hence $f(z)=\dfrac{e^{iz}}{(z^2+4)^2(z^2+9)}$" and proceed to calculate the contour integral of $f(z)$ without ever explaining why this is going to equal the integral with $\cos(z)$. Can anyone help me figure out why this is allowed?

Answer 1

$$e^{iz}=\cos{z}+i\sin{z}$$ Most of the time for contour integration if $f(x)$ has $\sin{x}$ OR $\cos{x}$ (as does this case), then you can express $\sin{x}$ as the imaginary part of $e^{ix}$ OR $\cos{x}$ as the real part of $e^{ix}$. Then, once you evaluate the contour integral, you take the real part of the answer if you let $\cos{x}=e^{ix}$ or take the imaginary part of the answer if you let $\sin{x}=e^{ix}$.

Answer 2

As @Ty noted, the trick is to write $\cos x=\Re\exp ix$. One advantage of this is that we can consider the behaviour of $\exp iz$ as $\Im z\to\pm\infty$ to identify the poles relevant to an application of the residue theorem. It's worth discussing your example to illustrate this. In the calculation $$\begin{align}\int_{\mathbb{R}}\frac{\cos xdx}{\left(x^{2}+4\right)^{2}\left(x^{2}+9\right)}&=\Re\int_{\mathbb{R}}\frac{\exp ixdx}{\left(x^{2}+4\right)^{2}\left(x^{2}+9\right)}\\&=\Re\left[2\pi i\left(\lim_{z\to2i}\frac{d}{dz}\frac{\exp iz}{\left(z+2i\right)^{2}\left(z^{2}+9\right)}+\lim_{z\to3i}\frac{\exp iz}{\left(z^{2}+4\right)^{2}\left(z+3i\right)}\right)\right],\end{align}$$the second $=$ includes the residues of the second-order pole $2i$ (see here) and the first-order pole $3i$. The integrand's other poles have negative imaginary part, so aren't enclosed in the "infinite semicircular contour" on the upper half of the complex plane. You can place a contour around them in the lower half, but therein $\exp iz$ diverges as $\Im z\to-\infty$, so these poles don't contribute to the integral at all.

I'll leave it to you to calculate the result now.