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

Consider \[ F(z)=\frac{z e^{iz}}{z^2+a^2},\qquad a>0, \] and integrate over the upper semicircle. By Jordan’s lemma the semicircular contribution tends to \(0\) as the radius tends to infinity. The only pole in the upper half-plane is the simple pole \(z=ia\), and \[ \operatorname{Res}_{z=ia}F(z) =\frac{ia\,e^{-a}}{2ia} =\frac{e^{-a}}2. \] Hence \[ \int_{-\infty}^{\infty}\frac{x e^{ix}}{x^2+a^2}\,dx =2\pi i\cdot\frac{e^{-a}}2 =\pi i e^{-a}. \] Taking imaginary parts gives \[ \boxed{\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+a^2}\,dx=\pi e^{-a}}. \]