I am able to derive the following equation by substituting the definition of a Fourier transform into it's inverse.

$$2\pi\delta(x-x') = \int_{-\infty}^{\infty} e^{ik(x-x')} dk$$

How do you prove that the Dirac Delta is equal to an integral of the exponential function? How do you prove the above equation is true?

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    $\begingroup$ Two answer this question, you need to specify the theoretical framework because this is not standard calculus. $\endgroup$ – Yves Daoust Aug 24 '17 at 19:07
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    $\begingroup$ @YvesDaoust The framework is the Fourier analysis where it is natural to apply the inversion theorem to the constant function $=1$, which leads directly to the distribution theory. $\endgroup$ – reuns Aug 24 '17 at 19:11
  • $\begingroup$ @reuns: I need the OP's statement. $\endgroup$ – Yves Daoust Aug 24 '17 at 19:13
  • $\begingroup$ You can't except someone who doesn't know how to prove the FIT for distributions to explain how to make this rigorous.. $\endgroup$ – reuns Aug 24 '17 at 19:14
  • $\begingroup$ For an elementary treatment see Lighthill's book on Fourier Analysis and Generalised Functions amazon.com/… $\endgroup$ – Spencer Aug 24 '17 at 19:21

Let $$h_a(x)= \int_{-a}^a e^{i k x} dk = \frac{2 \sin(a x)}{x}= a \, H'(ax), \\ H(x) = \int_{-\infty}^x \frac{2 \sin(y)}{y}dy, \qquad H(-\infty) = 0, H(+\infty) = C$$ where for some reason $C = 2\pi$

If $\phi,\phi'$ are $L^1$ then $$\lim_{a \to \infty}\int_{-\infty}^\infty h_a(x) \phi(x) dx = -\lim_{a \to \infty}\int_{-\infty}^\infty H(ax) \phi'(x) dx\\ = -\int_{-\infty}^\infty H(+\infty x) \phi'(x) dx = -\int_0^\infty C \phi'(x) dx= 2\pi \phi(0)$$ ie. in the sense of distributions $$\int_{-\infty}^\infty e^{ik x}dk \overset{def}=\lim_{a \to\infty} h_a = 2\pi \delta$$

Note how this proves the Fourier inversion theorem.

  • $\begingroup$ From the Dirichlet integrals it can be proven that the following is true. $$\int_{-\infty}^{\infty} \frac{2\sin(x)}x dx = 2\pi$$ Can you clarify what you mean by $\phi$ and $\phi'$ in the Lebesgue space of $L^1$? Is $\phi'$ the complex conjugate, implicit derivative or fourier transform of $\phi$? $\endgroup$ – Jeremy Aug 29 '17 at 2:22
  • $\begingroup$ @Jeremy I used integration by parts. And as my answer proves the Fourier inversion theorem, you can obtain $C = 2\pi$ from it. $\endgroup$ – reuns Aug 29 '17 at 11:10
  • $\begingroup$ The formula for integration by parts is $$(1) \space\space\space\space \int _{-\infty}^{\infty} udv \, = uv - \int_{-\infty}^{\infty} vdu\, dx $$ How did you assign the variables u and v? Suppose $u= H(ax)$ and $v = \phi(x)$ $$ \int_{-\infty}^{\infty} h_a(x) \phi(x) dx = - \int_{-\infty}^{\infty} H(ax) dx + (H(ax) \, \phi(x)) $$ In your equation you are missing the uv part of eq. 1. The limits have been removed for readability. $\endgroup$ – Jeremy Sep 1 '17 at 0:01
  • $\begingroup$ @Jeremy The assumption $\phi,\phi' \in L^1$ implies $uv|_{-\infty}^\infty = 0$. This is the same core idea as test functions allowing to differentiate distributions $\endgroup$ – reuns Sep 1 '17 at 0:07
  • $\begingroup$ What rule of distributions and the assumption of $\phi, \phi' \in L^1 $implies $uv|_{-\infty}^\infty = 0$? $\endgroup$ – Jeremy Sep 1 '17 at 0:30

We can give a meaning to $\int_{-\infty}^{\infty} e^{ikx} \, dk$ by introducing a damping factor $e^{-\frac12\epsilon k^2}$ inside the integral and at the end let $\epsilon \to 0$: $$ \int_{-\infty}^{\infty} e^{-\frac12\epsilon k^2} e^{ikx} \, dk = \int_{-\infty}^{\infty} e^{-\frac12\epsilon (k-ix/\epsilon)^2} e^{-\frac12 x^2/\epsilon} \, dk \\ = e^{-\frac12 x^2/\epsilon} \int_{-\infty}^{\infty} e^{-\frac12\epsilon (k-ix/\epsilon)^2} \, dk = \sqrt{\frac{2\pi}{\epsilon}} \, e^{-\frac12 x^2/\epsilon} \to 2\pi \, \delta(x) $$

  • $\begingroup$ As your post states, assume the following two equations are equal. $$\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\frac12\epsilon k^2} e^{ikx} \, dk $$ $$\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\frac12\epsilon (k^2-ix/\epsilon)^2} e^{-\frac12 x^2/\epsilon} \, dk$$ If both exponentials are continuous well-formed functions then the equations within the integrals must be equal. By the product rule of exponentials $$-\frac12\epsilon k^2 + ikx = -\frac12 \epsilon(k^2 - \frac{ix}\epsilon)^2 -\frac12 \frac{x^2}\epsilon$$ $\endgroup$ – Jeremy Aug 29 '17 at 3:39
  • $\begingroup$ Can you please explain the first equality? How did you move from eq. 1 to eq. 2? $\endgroup$ – Jeremy Aug 29 '17 at 3:54
  • $\begingroup$ Completion of the square: $$\frac12 \epsilon k^2 + ikx = \frac12 \epsilon \left( k^2 + 2\frac{ix}{\epsilon}k \right) = \frac12 \epsilon \left( k + \frac{ix}{\epsilon} \right)^2 - \frac12 \epsilon \left( \frac{ix}{\epsilon} \right)^2$$ Now, $$\frac12 \epsilon \left( \frac{ix}{\epsilon} \right)^2 = - \frac12 \epsilon \frac{x^2}{\epsilon^2} = -\frac{x^2}{2\epsilon}$$ $\endgroup$ – md2perpe Aug 29 '17 at 5:10
  • $\begingroup$ @md2perpe You need some complex analysis to justify the completion of the square. And it is not clear how you justify (in an elementary way) $\int_{-\infty}^\infty e^{ikx}dk = \lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\frac12\epsilon k^2} e^{ikx} \, dk$ $\endgroup$ – reuns Aug 29 '17 at 11:12
  • $\begingroup$ Of course you are correct, @reuns. The completion of the square itself is easily justified by basic complex algebra. That the integral can be translated in the imaginary direction is easily verified by a contour integral; the integrals over $[R, R+ix/\epsilon]$ and $[-R+ix/\epsilon, -R]$ tend to $0$ very fast as $R \to \infty$. $\endgroup$ – md2perpe Aug 29 '17 at 20:52

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