Clarification question about a common integral evaluated using residue theorem

Question

In the evaluation of $$\int_0^\infty \frac{\cos x}{1+x^2}\,dx\,,$$ you are to consider $$\frac{1}{2}\Re{\oint_C \frac{e^{iz}}{1+z^2}}\,dz\,,$$ where $C$ is the usual semi-circular $R\rightarrow \infty$ contour in the upper half-plane. My question is why would you not just consider $$\oint_C \frac{\cos z}{1+z^2}\,dz\,?$$ If you do, you get $$\frac{\pi}{2}\left(e+e^{-1}\right),$$ instead of the correct $$\frac{\pi}{2e}\,.$$ Why is that? I expect this to be a duplicate as I have seen two other people ask this question on another website, both with no response. If so, sorry.

Answer 1

If you use $\cos z$ then the intergal over the semi-circular portion does not tend to $0$ as $R \to \infty$. You do not have control over $|\cos (Re^{it})|$.

[ Note that $e^{-R \sin t}$ is bounded by $1$ but $e^{R \sin t}$ is not bounded].