I would like to show that $\int_{-\infty}^\infty {\sin(ax)\exp(ibx)\over x}dx$ is equals to $\pi$ for $b\in (-a,a)$ and equals to $0$ otherwise.
So I thought I shall use the keyhole contour integration, with a positively orientated contour in the upper half plane and having an indentation at $z=0$. Since $z=0$ is a simple pole, I can apply the indentation theorem which gives me (I believe) an arc integral of the indentation as $0$ and since my contour does not enclose any singularities, Cauchy's theorem suggests that the contour integral is $0$. Now I need to find the integral of the larger arc (of radius $R$, say) of the contour and find its limit as $R\to \infty$. But I don't quite know how.