Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 15A (Methods)

For every test function $f$, the change of variables $u=\alpha x$ givesEquality of distributions therefore gives

Using the definition of a distributional derivative,Hence $x{\delta }^{\prime }(x)=-\delta (x)$ as distributions.

Repeated integration by parts, or the definition of the $m$th distributional derivative, gives

Interpreting the displayed function aswe have ${\gamma }_n\ge 0$ and, after $u=nx$, ${\int }_{\mathbb{R}}{\gamma }_n=1$. Moreover its mass outside any fixed neighborhood of zero tends to zero. Therefore $({\gamma }_n)$ is an approximate identity, and for every test function $f$,Thus ${\gamma }_n\to \delta$ in the sense of distributions.

For a test function $f$, substitute $y=nx$ and use the Taylor expansion $f(y/n)=f(0)+(y/n)f^{\prime }(0)+O(n^{-2})$. Oddness removes the first term, while the Gaussian integral gives $\int y^2{e}^{-y^2}dy=\sqrt{\pi }/2$. HenceTherefore

Direct integration givesThe sinc approximate identity converges distributionally to the Dirac delta function, using the supplied integral of $\sin y/y$. Thus the limit is $\delta (x)$.

Since the odd sine part of the complex exponential integrates to zero,Part (e)(ii) therefore yieldsin the sense of distributions. This is the Fourier representation of the Dirac delta function.