the large semicircle contour

To turn a real integral over the whole line into a closed loop the residue theorem can attack, you need to bend the line back on itself without changing its value. The large semicircle is the standard way: keep the real segment from -R to R, then return from R to -R along a big half-circle arc that bulges into the upper (or lower) half-plane. The result is a D-shaped closed contour.

Concretely the contour has two pieces: the diameter, which is the real interval [-R, R] traversed left to right, and the arc C_R, the upper semicircle |z| = R run counterclockwise from R back to -R. Apply the residue theorem to the whole loop: the integral over the segment plus the integral over C_R equals 2 pi i times the residues enclosed. The strategy is then to let R go to infinity. The segment integral becomes the real integral you actually want. The whole method succeeds or fails on one question: does the integral over the arc C_R go to zero as R grows? If yes, the real integral equals the residue sum.

Whether the arc vanishes depends on how fast the integrand decays. For a rational function P/Q with deg Q at least deg P + 2, the ML inequality settles it: on the arc |P/Q| is at most roughly 1/R^2 and the arc length is pi R, so the arc integral is bounded by something like pi R / R^2 = pi / R, which goes to 0. For oscillatory integrands like e^(i a x) f(x) with slower decay, the crude ML bound is not enough and you upgrade to Jordan's lemma, which exploits the exponential's decay in the upper half-plane. Choosing upper versus lower half-plane is dictated by where the integrand decays — for e^(i a x) with a positive, the factor e^(i a z) decays in the upper half-plane, so you close upward.