 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/7d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

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

 New to topics? Read the docs here!