 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/7a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Apply the residue theorem to the upper-half-plane semicircle, with a small semicircular indentation above the pole at

z = 0

. The large arc contributes zero because

f (z) / z \to 0

, so

\frac{f ( z )}{z ( z ^{2} + a ^{2} )} = o (∣ z ∣^{- 2}) .

The small indentation is traversed clockwise and contributes

- iπ

times the residue at zero. The only pole strictly inside the indented contour is

z = ia

. Therefore

PV \int_{- \infty}^{\infty} \frac{f ( x )}{x ( x ^{2} + a ^{2} )} d x = 2 πi Res_{z = ia} \frac{f ( z )}{z ( z ^{2} + a ^{2} )} + iπ Res_{z = 0} \frac{f ( z )}{z ( z ^{2} + a ^{2} )} .

The two residues are

Res_{z = ia} = - \frac{f ( ia )}{2 a ^{2}}, Res_{z = 0} = \frac{f ( 0 )}{a ^{2}} .

PV \int_{- \infty}^{\infty} \frac{f ( x )}{x ( x ^{2} + a ^{2} )} d x = \frac{iπ}{a ^{2}} (f (0) - f (ia)) .

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