Evaluate:$$I=\int^{\infty}_{-\infty}\frac{\cos xdx}{x+i}$$
which means that $$f(z) = \frac{e^{iz}}{z+i}$$ with the simple pole $z=-i$ and since $z=-i$ i'd have to integrate it using the bottom half of the semi-circle.
$$\Longrightarrow \int_C f(z)dz=\int_\Gamma+\int^R_{-R} f(x)dx=2\pi iRes_{z\to-i}\Big(z-(-i)\Big)\frac{e^{iz}}{z+i}=e$$
$$\int_{\Gamma}f(z)dz=\int^{-R}_{R}\frac{e^{iz}}{z+i}dz$$
I'm not sure on what to do next.
Use the top half circle $C_+$ centered at $0$ with radius $R$ to compute$$\int_{C_+}\frac{e^{iz}}{z+i}\,\mathrm dz.$$This integral is $0$; of course, and therefore$$\int_{-\infty}^\infty\frac{e^{iz}}{z+i}\,\mathrm dz=0.$$Now, consider the bottom half-circle $C_-$ centered at $0$ with radius $R$. If $R>1$, then$$\int_{C_-}\frac{e^{-iz}}{z+i}=-\frac{2\pi i}e$$and therefore$$\int_{-\infty}^\infty\frac{e^{-iz}}{z+i}\,\mathrm dz=-\frac{2\pi i}e.$$And now use this to compute$$\int_{-\infty}^\infty\frac{e^{ix}+e^{-ix}}{2(x+i)}\,\mathrm dx=\int_{-\infty}^\infty\frac{\cos x}{x+i}\,\mathrm dx.$$