Discussion
Real integrals over \( (-\infty, \infty) \) are evaluated by residues using a contour that is
Answer: D
The real line is closed by a semicircle in the upper half plane; if the integrand decays fast enough, that arc contributes nothing in the limit and the real integral equals \( 2\pi i \) times the residues in the upper half plane. A small circle round the origin is the tool for a principal value, not for the whole line
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