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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!

Suppenkasper
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2 Answers2

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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 $u = \epsilon x$.

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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.

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

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

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

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

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!

Mark Viola
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  • I like your answer! Since you answered after Open Ball I am wondering if it is necessary to take the extra steps that you took? If so what is it that you have to be aware of? – Suppenkasper Jan 22 '17 at 20:03
  • @Suppenkasper Well, I didn't use the approach that open ball suggested. Note that I didn't have to split the integration into two parts. That said, there is always more than one way to proceed. ;-)) – Mark Viola Jan 22 '17 at 20:18
  • I upvoted both your answers but its not visible since i do not have enough reputation yet ;) I will best vote @Open-Ball's answer for its simplicity. Thanks a lot though! – Suppenkasper Jan 22 '17 at 23:26
  • @Suppenkasper How do you proceed with open-ball's hint? You need to justify that the second integral goes to $0$. My solution shows how to do this. – Mark Viola Jan 23 '17 at 00:02
  • Sorry I think I might not understand what you are getting at. Why is it necessary to show that $| \delta_{\epsilon} (x)|$ is not integrable? – Suppenkasper Jan 23 '17 at 09:51
  • It's not necessary to show that $|\delta_\epsilon|$ isn't integrable. $|\delta_\epsilon|=\sqrt{a/\pi}$, assuming of course that $a$ is real and positive. And the fact that it isn't integrable poses a bit of a challenge to showing that $\delta_\epsilon $ is indeed a regularization of the Dirac Delta. – Mark Viola Jan 23 '17 at 15:27