Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 14D (Methods)

Use the Fourier transform conventionDecay alone does not guarantee existence of the ordinary integrals, so impose the usual integrability and regularity assumptions. One sufficient version of the Fourier inversion theorem is that $f$ and $\tilde{f}$ are integrable and $f$ is continuous, in which case the inversion integral recovers $f$. Under the standard piecewise-smooth Dirichlet assumptions, inversion instead gives the midpoint of the two one-sided limits at a jump, with a suitable improper integral interpretation.

For integrable $f,g$, define $(f\ast g)(x)={\int }_{\mathbb{R}}f(x-y)g(y)dy$. The convolution theorem statesTo prove it, absolute integrability permits Fubini's theorem. Set $t=x-y$ in the inner integral:The absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$, justifying the interchange. The computations below assume $a,b>0$ and use the source PDF's factor $e^{ikx}$ in the final integral, omitted in the converted TeX.

Split the exponential kernel at zero. For $a>0$,Thus its Fourier transform isBoth half-line integrals converge absolutely because $a>0$.

The interval indicator has transformwith continuous value $2b$ at $k=0$. By the convolution theorem, $h=f\ast g$ has transform $4a\sin (bk)/[k(a^2+k^2)]$. Fourier inversion therefore expresses the requested integral asFor $x>b$, $|x-y|=x-y$, so $h(x)=2e^{-ax}\sinh (ab)/a$. Symmetry gives the same expression with $|x|$ for $x<-b$. For $|x|\le b$, splitting the integral at $y=x$ givesConsequently the value for every real $x$ isThe two expressions agree at $x=\pm b$. This is the exponential kernel convolved with an interval indicator. The spectral integral is absolutely convergent: its apparent singularity at zero is removable, and the tail decays as $O(|k{|}^{-3})$.