Exercise · ZW-REAL-INTEGRAL-BY-RESIDUES-009 · ★★★ Advanced

Let \(f=P/Q\) be rational with no real poles. Which condition guarantees that the integral over the semicircle \(C_R\) tends to \(0\) as \(R\to\infty\)?

Choose one option.

Show answer
  • A. \(\deg Q\ge \deg P+2\)

Explanation

If \(\deg Q\ge\deg P+2\), then \(|f(z)|\le M/R^2\) on \(C_R\) and the integral is bounded by \(M\pi R/R^2\to0\). With difference \(1\) one only gets \(|f|\sim 1/R\) and the bound does not vanish.

Study Complex analysis in ZWMath →

Review the concept: Real integral by residues

A method to compute \(\int_{-\infty}^{\infty} f(x)\,dx\) with \(f\) rational: close the real axis with a semicircle in the upper half-plane and apply the residue theorem.

\[\int_{-\infty}^{\infty} f(x)\,dx=2\pi i\sum_{\operatorname{Im}z_k>0}\operatorname{Res}(f,z_k)\]

More practice problems: real integral by residues

← ZW-REAL-INTEGRAL-BY-RESIDUES-008 ZW-REAL-INTEGRAL-BY-RESIDUES-010 →