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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 7D / Solution

The ordinary integral fails because the simple real poles at

- 2

and

3

produce nonintegrable logarithmic singularities. Partial fractions give

\frac{1}{x ^{2} - x - 6} = \frac{1}{5} (\frac{1}{x - 3} - \frac{1}{x + 2}) .

Closing above for

ω > 0

and below for

ω < 0

, with symmetric semicircular indentations, yields

I (ω) = \frac{iπ}{5} sgn (ω) (e^{3 iω} - e^{- 2 iω}) .

Taking real and imaginary parts,

PV \int \frac{c o s ω x}{x ^{2} - x - 6} d x = - \frac{π}{5} sgn (ω) (sin 3 ω + sin 2 ω),

PV \int \frac{s i n ω x}{x ^{2} - x - 6} d x = \frac{π}{5} sgn (ω) (cos 3 ω - cos 2 ω) .

All integrals here run over

R

.

Solved by gpt-5.6-sol high.

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