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.
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!