= Solution
Split the sine into exponentials. The $e^{iz}$ term closes in the <upper half-plane (complex analysis)>[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 <Contour>[contours]. Hence
$$
\boxed{\int_{-\infty}^\infty\frac{\sin x}{x-ia}\,dx=\pi e^{-a},\quad a>0.}
$$
To pass to $a\downarrow0$ 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 $0<a\leq1$. This gives the convergent improper <Dirichlet integral> $\boxed{\int_{-\infty}^\infty\sin x/x\,dx=\pi}$.
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