Evaluating real integrals through residues

Question

Say we are trying to evaluate a real integral of the form $$\int_{-\infty}^\infty f(x)\,dx$$ through the use of residues. Typically, the contour used is a closed semi-circle formed by the segment that goes from $(-R,0)$ to $(R,0)$ and $C_R: |z|=R$. For this question, we can assume that $f(z)$ has a finite number of singularities inside the contour, then, the approach would be to evaluate $$\int_{-R}^Rf(x)\,dx=2\pi i \sum_{k=0}^n\text{Res}(f,z_k)-\int_{C_R}f(z)\,dz$$ and then show that the integral on the right of the equality approaches zero through the Darboux inequality (then of course take the real part and so on). However, my question is, if we know that the initial (real) integral converges, do we have to show that such integral approaches zero? In other words, is it enough to write it as $$\int_{-\infty}^{\infty} f(x)\,dx=\Re\left[2\pi i \sum_{k=0}^n\text{Res}(f,z_k)\right]$$

Answer 1

As an example, consider $$ I = \int_{-\infty}^{\infty} \frac{\cos x}{x^2+1}\,dx $$ and choose $f(z) = \dfrac{\cos z}{z^2+1}$ (normally you would take $f(z) = \frac{e^{iz}}{z^2+1}$ to compute $I$ via residues). Cleary the integral converges and you can check that $I = \dfrac{\pi}{e}$, but $$ 2\pi i\operatorname{Res}_{z=i} \frac{\cos z}{z^2+1} = \pi\cosh 1. $$