 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2/2/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Use the complex exponential representation of cosine. For real

α

and

β > 0

, exponential decay makes the improper integral absolutely convergent, and

I (α, β) = Re \int_{0}^{\infty} e^{- (β - i α) x} d x = Re \frac{1}{β - i α} = \frac{β}{β ^{2} + α ^{2}} .

β

. The minus sign is essential:

\partial_{β} e^{- β x} = - x e^{- β x}

. Hence

\int_{0}^{\infty} x cos (αx) e^{- β x} d x = - \frac{\partial I}{\partial β} = \frac{β ^{2} - α ^{2}}{( β ^{2} + α ^{2} ) ^{2}} .

To justify this step near any fixed

β_{0} > 0

, restrict

β \geq β_{0} /2

. The absolute value of the differentiated integrand is at most

x e^{- β_{0} x /2}

, an integrable bound, so dominated convergence permits the differentiation. At

α = 0

the answers reduce to

1/ β

and

1/ β^{2}

, respectively.

 New to topics? Read the docs here!