Fresnel function converging to delta distribution

Question

i need to show that the function known from the Fresnel integral (wikipedia) converges to the Dirac delta-distribution. This function is defined as

$f_{\epsilon}(x) = \sqrt{\frac{a}{i \pi}\frac{1}{\epsilon}} e^{ia\frac{x^2}{\epsilon}} $.

More concretely this means that i need to show that

$\lim_{\epsilon \rightarrow 0} \; \sqrt{\frac{a}{i \pi}\frac{1}{\epsilon}} e^{ia\frac{x^2}{\epsilon}} = \delta(x)$.

I allready know that one criterium for this to be the delta-distribution namely normalization is fullfilled: $\int_{\mathbb{R}} \sqrt{\frac{a}{i \pi}\frac{1}{\epsilon}} e^{ia\frac{x^2}{\epsilon}}dx = 1$.

What i do not know how to show is that this:

$\lim_{\epsilon \rightarrow 0}\int_{\mathbb{R}} f_{\epsilon} (x) g(x) dx = g(0) $

Any help is much appreciated!

Answer 1

Hint: write

$$\int f_{\epsilon} g = g(0) + \int_{\Bbb R} f_{\epsilon}(x) (g(x) - g(0)) dx$$

to show that the integral $\to 0$, apply the change of variables $u = \epsilon x$.

Answer 2

To show that $\delta_{\epsilon}(x)= \sqrt{\frac{a}{i\pi}}\frac1\epsilon e^{iax^2/\epsilon}$ is a regularization of the Dirac Delta distribution, we need to show that

$$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \phi(x)\delta_{\epsilon}(x)\,dx=\phi(0)$$

for any suitable test function. One challenge in showing this is that $|\delta_{\epsilon}(x)|$ is not integrable. This challenge will met by a simple contour deformation followed by applying the Dominated Convergence Theorem.

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Let $\phi(x)$ be a suitable test function. Enforcing the substitution $x\to \sqrt{\epsilon/a}\,x$, we have

$$\begin{align} \lim_{\epsilon\to 0}\int_{-\infty}^\infty \phi(x)\delta_{\epsilon}(x)\,dx&=\lim_{\epsilon\to 0}\frac1\epsilon\sqrt{\frac{a}{i\pi}}\int_{-\infty}^\infty\phi(x)e^{iax^2/\epsilon}\,dx\\\\ &=2\sqrt{\frac{1}{i\pi}}\lim_{\epsilon\to 0}\int_{0}^\infty \phi\left(\sqrt{\epsilon/a}\,x\right)e^{ix^2}\,dx \end{align}$$

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Contour Deformation:

Next, using Cauchy's Integral Theorem, we deform the contour from the real line to the ray $z=xe^{i\pi/4}$ to find

$$\begin{align} \lim_{\epsilon\to 0}\int_{-\infty}^\infty \phi(x)\delta_{\epsilon}(x)\,dx&=2\sqrt{\frac{1}{i\pi}}\lim_{\epsilon\to 0}\int_{0}^\infty \phi\left(\sqrt{\epsilon/a}\,xe^{i\pi/4}\right)e^{-x^2}\,e^{i\pi/4}\,dx\\\\ &=2\sqrt{\frac{1}{\pi}}\lim_{\epsilon\to 0}\int_{0}^\infty \phi\left(\sqrt{\epsilon/a}\,xe^{i\pi/4}\right)e^{-x^2}\,dx\\\\ \end{align}$$

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Applying the Dominated Convergence Theorem:

Invoking the Dominated Convergence Theorem reveals

$$\begin{align} \lim_{\epsilon\to 0}\int_{-\infty}^\infty \phi(x)\delta_{\epsilon}(x)\,dx& =2\sqrt{\frac{1}{\pi}}\lim_{\epsilon\to 0}\int_{0}^\infty \phi\left(\sqrt{\epsilon/a}\,xe^{i\pi/4}\right)e^{-x^2}\,dx\\\\ &=2\sqrt{\frac{1}{\pi}}\int_{0}^\infty \lim_{\epsilon\to 0}\phi\left(\sqrt{\epsilon/a}\,xe^{i\pi/4}\right)e^{-x^2}\,dx\\\\ &=2\sqrt{\frac{1}{\pi}}\phi(0)\int_0^\infty e^{-x^2}\,dx\\\\ &=\phi(0) \end{align}$$

as was to be shown!