Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 3

Use the Fourier transform conventionThese hold directly under suitable integrability and inversion hypotheses, and extend to square-integrable functions through the Plancherel theorem. At a jump, ordinary Fourier inversion uses the mean of the one-sided limits under the usual piecewise smooth hypotheses.

For the two-sided exponential, split the integral at zero:For the interval indicator, the Fourier transform of an interval indicator givesAt $k=0$, the continuous limiting value is $4\pi$, the length of the interval. Changing the values at its two endpoints has no effect on the transform.

The convolution is $(u\ast v)(x)={\int }_{-\infty }^{\infty }u(t)v(x-t)dt$. In this transform convention, the convolution theorem says $\widehat{u\ast v}=\hat{u}\hat{v}$, and Parseval's identity saysTo match the integral's scale exactly, set $u=f/2$ and $h(x)=\frac{1}{2}{1}_{[-1,1]}(x)=g(2\pi x)/2$. Then $\hat{u}=1/(1+k^2)$ and $\hat{h}=\sin k/k$, using the value one at zero. Their convolution is the box smoothing of a two-sided exponential with both scales equal to one:Its transform is precisely $\sin k/[k(1+k^2)]$. Since $w$ is even, Parseval's identity reduces the requested integral toUse ${\int }_0^1\cosh xdx=\sinh 1$, ${\int }_0^1{\cosh }^{2}xdx=1/2+\sinh 2/4$ and ${\int }_1^{\infty }{e}^{-2x}dx=e^{-2}/2$. Simplifying givesThe apparent singularity at $k=0$ is removable, and the integrand decays as $O(|k{|}^{-6})$, so the integral converges.