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

Use the complex exponential representation of cosine. For real $\alpha$ and $\beta >0$, exponential decay makes the improper integral absolutely convergent, andFor the second integral, use differentiation under the integral sign with respect to $\beta$. The minus sign is essential: ${\partial }_{\beta }{e}^{-\beta x}=-x{e}^{-\beta x}$. HenceTo justify this step near any fixed ${\beta }_0>0$, restrict $\beta \ge {\beta }_0/2$. The absolute value of the differentiated integrand is at most $x{e}^{-{\beta }_{0}x/2}$, an integrable bound, so dominated convergence permits the differentiation. At $\alpha =0$ the answers reduce to $1/\beta$ and $1/{\beta }^2$, respectively.