$\int_{-\infty}^\infty\frac{\cos x}{x^2+a^2}\,dx=\pi\frac{e^{-a}}{a}$

Consider \[ F(z)=\frac{e^{iz}}{z^2+a^2},\qquad a>0, \] and integrate over the upper semicircle of radius \(R>a\). Since \(|e^{iz}|=e^{-\Im z}\le1\) there and \(|z^2+a^2|\ge R^2-a^2\), the arc integral tends to \(0\) as \(R\to\infty\).

The only pole in the upper half-plane is the simple pole \(z=ia\), with residue \[ \operatorname{Res}_{z=ia}F(z) =\frac{e^{i(ia)}}{2ia} =\frac{e^{-a}}{2ia}. \] Thus the residue theorem gives \[ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^2+a^2}\,dx =2\pi i\frac{e^{-a}}{2ia} =\frac{\pi e^{-a}}a. \] Taking real parts yields \[ \boxed{\int_{-\infty}^{\infty}\frac{\cos x}{x^2+a^2}\,dx =\pi\frac{e^{-a}}a}. \]