Proving $\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk=\delta(x-a)$ with this method I want to prove $$\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk=\delta(x-a)$$
By following the following logic:
$$\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk$$
equals $0$ whenever $x\ne a$ and $\infty$ when $x=a$
So the next step would be to prove that
$$\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk dx=1$$
I've tried to begin by separating it as
$$\int^{\infty}_{-\infty}e^{2\pi i k a}\int^{\infty}_{-\infty}e^{-2\pi i k x}dx dk$$
But I am stuck. Please help.
 A: We need to show that for any test function $\phi(x)$ that 
$$\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x)\int_{-L}^L e^{-i2\pi k(x-a)}\,dk\,dx=\phi(a)$$
Proceeding, we have for any $\epsilon>0$
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x)\int_{-L}^L e^{-i2\pi k(x-a)}\,dk\,dx&=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x)\left(\frac{\sin(2\pi (x-a)L)}{\pi (x-a)}\right)\,dx\\\\
&=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x+a)\left(\frac{\sin(2\pi xL)}{\pi x}\right)\,dx\\\\
&=\lim_{L\to\infty}\left(\int_{|x|\le \epsilon}\phi(x+a)\left(\frac{\sin(2\pi xL)}{\pi x}\right)\,dx\right.\\\\
&+\left.\int_{|x|\ge \epsilon}\phi(x+a)\left(\frac{\sin(2\pi xL)}{\pi x}\right)\,dx\right)\tag1\\\\
&=\lim_{L\to\infty}\int_{|x|\le L\epsilon}\phi(x/L+a)\left(\frac{\sin(2\pi x)}{\pi x}\right)\,dx\tag2\\\\
&=\lim_{L\to\infty}\int_{|x|\le L\epsilon}\left(\phi(a)+O\left(\frac xL\right)\right)\left(\frac{\sin(2\pi x)}{\pi x}\right)\,dx\\\\
&=\phi(a)+O(\epsilon)\tag3
\end{align}$$
In going from $(1)$ to $(2)$, we applied the Riemann-Lebesgue Lemma.
Finally, since, $\epsilon>0$ is arbitrary, we may take the limit as $\epsilon\to 0$ of $(3)$ to find that 
$$\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x)\int_{-L}^L e^{-i2\pi k(x-a)}\,dk\,dx=\phi(a)$$
as was to be shown!
