an indented contour
The residue theorem demands that singularities lie strictly inside the contour, never on it. But many real integrals — especially Fourier-type ones — have a pole sitting right on the real axis, exactly on the path you want to integrate along. An indented contour is the fix: you make a tiny detour, a small semicircle, that steps around the offending pole, keeping the path away from the singularity while changing the value by only a controlled, computable amount.
How it works for a simple pole at x = c on the real axis. Instead of running straight through c, the contour dips below (or arcs above) with a small semicircle of radius epsilon centered at c. As epsilon shrinks to zero, two things happen: the straight pieces on either side of c combine into the principal value of the real integral, and the small semicircle contributes a precise, non-vanishing amount. The key fact: the integral over a small semicircle of radius epsilon around a simple pole, as epsilon goes to 0, tends to plus or minus pi i times the residue at that pole — note the half, pi i not 2 pi i, because the semicircle subtends an angle of only pi rather than the full 2 pi. The sign depends on whether the detour goes clockwise (minus) or counterclockwise (plus).
Putting it together with the residue theorem yields the working formula for a pole on the real axis: the principal value of the real integral equals 2 pi i times the residues strictly enclosed, minus the pi i times residue contributed by each indented real pole (with sign set by how you indented). This is the standard route to integrals like the integral of (sin x) / x dx, whose integrand has a pole at the origin. The crucial honesty: the small arc does NOT vanish — it contributes exactly a half-residue, and forgetting that half is one of the most common errors in contour integration.
For the integral over the real line of (sin x)/x dx, take e^(i z)/z with a small semicircle indenting above the pole at 0. The big arc dies by Jordan's lemma, the loop encloses no pole, and the small arc gives -pi i times Res = -pi i; balancing yields the principal value pi.
A small semicircle around a simple real pole contributes exactly pi i times the residue, with a sign.
The half-residue rule pi i times the residue holds only for a simple pole; for a higher-order pole the small-arc integral does not tend to a finite multiple of the residue and the method must be reworked. Track the indentation's orientation carefully — it sets the sign.