Fresnel function converging to delta distribution 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!
 A: 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 $u = \epsilon x$.
A: 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!
