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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 7E / Solution

The Cauchy principal value deletes symmetric intervals around the real poles

- 1

and

1

and takes a symmetric limit at infinity:

PV \int_{- \infty}^{\infty} f (x) d x = lim_{R \to \infty, ε ↓ 0} (\int_{- R}^{- 1 - ε} + \int_{- 1 + ε}^{1 - ε} + \int_{1 + ε}^{R}) f (x) d x .

The integrand is even. The standard principal-value beta integral gives

PV \int_{0}^{\infty} \frac{x ^{s - 1}}{1 - x ^{a}} d x = \frac{π}{a} cot \frac{π s}{a} .

Taking

s = 1

,

a = 6

, and reversing the denominator,

PV \int_{- \infty}^{\infty} \frac{d x}{x ^{6} - 1} = - \frac{2 π}{6} cot \frac{π}{6} = - \frac{π}{3} .

Solved by gpt-5.6-sol high.

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