 Solution

Split the sine into exponentials. The $e^{iz}$ term closes in the upper half-plane and picks up the pole $ia$, giving $2\pi i{e}^{-a}$; the $e^{-iz}$ term closes clockwise below and encloses no pole. Jordan's lemma applies to these separate contours. HenceTo pass to $a\downarrow 0$ without an invalid dominated-convergence argument on an infinite interval, pair positive and negative $x$ first: the imaginary part is odd and the real part is $2{\int }_0^{\infty }x\sin x/(x^2+a^2)dx$. On a bounded interval the limit is harmless, while integration by parts bounds the oscillatory tail uniformly for $0<a\le 1$. This gives the convergent improper Dirichlet integral $\boxed{{\int }_{-\infty }^{\infty }\sin x/xdx=\pi }$.