 Home

Codex Wiki

List of universities

University of Cambridge

Course of the University of Cambridge

Natural sciences course of the University of Cambridge

Year of the natural sciences course of the University of Cambridge

Year 2 of the natural sciences course of the University of Cambridge

Mathematics course of the year 2 of the natural sciences course of the University of Cambridge

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge

2024

Paper 1

The squared <norm> of the original function is
$$
\int_{\mathbb R}|f(x)|^2\,dx=\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\frac\pi2.
$$
Applying the <Parseval identity> to part (c), using the evenness of the squared transform, gives
$$
\frac\pi2=\frac1{2\pi}\int_{\mathbb R}\frac{4k^2\cos^2(\pi k/2)}{(k^2-1)^2}\,dk=\frac4\pi\int_0^{\infty}\frac{k^2\cos^2(\pi k/2)}{(k^2-1)^2}\,dk.
$$
Thus the last <integral> equals $\pi^2/8$. Now set $u=\pi k/2$. Since
$$
\frac{k^2\,dk}{(k^2-1)^2}=\frac\pi2\frac{u^2\,du}{(u^2-\pi^2/4)^2},
$$
we obtain
$$
\boxed{\int_0^{\infty}\frac{u^2\cos^2u}{(u^2-\pi^2/4)^2}\,du=\frac2\pi\frac{\pi^2}{8}=\frac\pi4.}
$$
The apparent singularity at $u=\pi/2$ is removable: the continuous value of the integrand there is $1/4$. No principal-value interpretation is needed.

Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 4 / d / Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 4 / d / Solution

 Ancestors (13)

 Discussion (0)

 Articles by others on the same topic (0)

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 4 / d / Solution

 Ancestors (13)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)