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.

  • 1
    $\begingroup$ and why would $$\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk dx=1$$ imply that $$\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk=\delta(x-a)$$ ???? $\endgroup$ – Ahmad Bazzi Sep 27 '18 at 17:09
  • $\begingroup$ Because the properties of the Dirac-Delta function are that $\delta (x-a)$ is zero everywhere except for $x=a$ where it has infinite value, and it's integral over all of $x$ equals one. Since $\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk$ is zero everywhere except when $x=a$, we only need to prove that it's integral is equal to one, and it'll have the same properties of the Dirac-Delta function, and therefore be equal to the Dirac-Delta function. $\endgroup$ – user140323 Sep 27 '18 at 17:49
  • $\begingroup$ Also, it is known that $$\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk=\delta(x-a)$$ from Fourier Analysis, I just want to prove it in a different way. $\endgroup$ – user140323 Sep 27 '18 at 17:50
  • $\begingroup$ This $\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk=\delta(x-a)$ is correct and here is one way of proving it: math.stackexchange.com/questions/2930585/…. But trying to prove it by proving $\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-2\pi i k (x-a)}dk dx=1$ is wrong. Do you see why ? $\endgroup$ – Ahmad Bazzi Sep 27 '18 at 17:51
  • $\begingroup$ I don't quite understand why. Care to elucidate? $\endgroup$ – user140323 Sep 27 '18 at 19:00

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!

  • $\begingroup$ Hi. Would you please let me know how I can improve my answer? I really want to give you the best answer I can. If this was not useful, I am happy to delete it. Looking forward to your reply. $\endgroup$ – Mark Viola Nov 17 '18 at 18:26
  • $\begingroup$ And feel free to up vote and accept an answer as you see fit. ;-) $\endgroup$ – Mark Viola Nov 17 '18 at 18:28

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