Keyhole contours evaluate Mellin-type rational integrals

Let ${\mathrm{\Gamma }}_{\epsilon ,\rho }$ be the keyhole contour of inner radius $\epsilon$ and outer radius $\rho$. Applying [L2] to $f(z)=z^{\alpha -1}R(z)$ on the slit annulus gives ${\int }_{{\mathrm{\Gamma }}_{\epsilon ,\rho }}f(z)dz=2\pi i\sum \mathrm{Res}(f,a),$ where the sum is over the enclosed poles away from the positive axis.

The contour integral splits into outer circle, upper lip, inner circle, and lower lip. By hypothesis the two circular contributions vanish. The assumed convergence of the improper integral and the upper boundary value in [L1] make the upper lip tend to ${\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}R(x)dx$. The lower lip has the reverse orientation and boundary value $e^{2\pi i\alpha }{x}^{\alpha -1}R(x)$, so it tends to $-e^{2\pi i\alpha }{\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}R(x)dx$.

Taking $\epsilon \downarrow 0$ and $\rho \to \mathrm{\infty }$ in step 1.1 and substituting the boundary terms from step 1.2 yields $(1-e^{2\pi i\alpha }){\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}R(x)dx=2\pi i\sum \mathrm{Res}(f,a)\mathrm{.}$ The finite set of poles of the rational factor is eventually enclosed, so this is the sum over all poles of $f$ away from the positive axis and hence the claimed keyhole identity.