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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?

Jeremy
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    Two answer this question, you need to specify the theoretical framework because this is not standard calculus. –  Aug 24 '17 at 19:07
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    @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. – reuns Aug 24 '17 at 19:11
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    @reuns: I need the OP's statement. –  Aug 24 '17 at 19:13
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    You can't except someone who doesn't know how to prove the FIT for distributions to explain how to make this rigorous.. – reuns Aug 24 '17 at 19:14
  • For an elementary treatment see Lighthill's book on Fourier Analysis and Generalised Functions https://www.amazon.com/Introduction-Generalised-Functions-Cambridge-Monographs/dp/0521091284/ref=sr_1_1?ie=UTF8&qid=1503602451&sr=8-1&keywords=lighthill+generalized+functions – Spencer Aug 24 '17 at 19:21
  • @reuns, my comment wasn't meant to denigrate your answer in any way. I would agree that your answer is elementary, and as far as I can tell it is a good answer. I only mentioned Lighthill's book to provide the OP with a good reference, since I don't have the time to write my own answer here. Incidentally Lighthill's proof of the Fourier Inversion theorem is slightly different than your own so its possible the OP might benefit from the comparison. – Spencer Aug 24 '17 at 19:31
  • The context I am using the dirac delta in is quantum mechanics. Expected momentum is defined as $$ (4)\space\space\space\space\space\space\space\space\space\space\space\space〈\space p \space 〉 =\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\Bigg(\int_{-\infty}^{\infty} \hat \phi\dot(j) \space e^{-ijx} \delta j\Bigg)\space\space \Bigg(\int_{-\infty}^{\infty} \hbar k\space \space \hat \phi(k) \space e^{ikx} \delta k\Bigg)\delta x $$ – Jeremy Aug 29 '17 at 03:59
  • Equation 4 simplifies to

    $$(5)\space\space\space\space\space\space\space\space\space\space\space\space〈\space p \space 〉 = \frac{1}{2\pi}\Bigg(\int_{-\infty}^{\infty} \int_{-\infty}^\infty \hat \phi\dot(j) \hat \phi(k) ,\hbar k\space \int_{-\infty}^{\infty} \space e^{i(k-j)x} \space\space \space \space \delta x , \delta j \delta k\Bigg)$$

     – Jeremy Aug 29 '17 at 04:14
  • Then, the dirac delta is introduced from $\int_{-\infty}^\infty e^{i(k-j)x} \delta x$ $$ (6)\space\space\space\space\space\space\space\space\space\space\space\space〈\space p \space 〉 = \Bigg(\int_{-\infty}^{\infty} \int_{-\infty}^\infty \hat \phi\dot(j) \hat \phi(k) ,\hbar k\space \delta (k-j), \delta j \delta k\Bigg)$$ – Jeremy Aug 29 '17 at 04:19

2 Answers2

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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.

reuns
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    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 $\phi$?

     – Jeremy Aug 29 '17 at 02:22
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    @Jeremy I used integration by parts. And as my answer proves the Fourier inversion theorem, you can obtain $C = 2\pi$ from it. – reuns Aug 29 '17 at 11:10
  • 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.

     – Jeremy Sep 01 '17 at 00:01
  • @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 – reuns Sep 01 '17 at 00:07
  • What rule of distributions and the assumption of $\phi, \phi' \in L^1 $implies $uv|_{-\infty}^\infty = 0$? – Jeremy Sep 01 '17 at 00:30
  • Also, how do you move from an integration with limits of $[-\infty, \infty]$ to $[0, \infty]$? – Jeremy Sep 01 '17 at 00:36
  • @Jeremy I made it clear : $$H(+\infty x) = \ ?$$ About integrating by parts, can you write how you would integrate by parts $\int_{-\infty}^\infty \log( x) e^{-x^2}dx$ ? – reuns Sep 01 '17 at 00:41
  • That is not related to distributions. I don't think there is a valid integration by parts method for that particular equation.

    I understand the following equality.

    $$H(+\infty,x),=, 2\pi$$

    I don't understand how the limits of integration changed from $[-\infty, \infty]$ to $[0, \infty]$.

    Also I don't know what rule of distributions implies that if two functions, $\phi$ and $\phi' \in L^1$ and $v=\phi$ then $u\phi|_{-\infty}^\infty = 0$.

     – Jeremy Sep 01 '17 at 01:03
  • @Jeremy Let $u(x) = e^{-x^2}, v'(x) = \log(x), u'(x) = -2x e^{-x^2}, v(x) = x (\log x-1)$. Then $$\int_{-\infty}^\infty \log( x) e^{-x^2}dx = \lim_{b \to \infty} \int_{-b}^b u(x) v'(x)dx = \lim_{b \to \infty} u(x)v(x)|{-b}^b - \int{-b}^b u'(x) v(x)dx \ = -\int_{-\infty}^\infty u'(x) v(x)dx= \int_{-\infty}^\infty 2x e^{-x^2} x (\log (x)-1)dx$$ where $ \lim_{b \to \infty} u(x)v(x)|_{-b}^b = 0$ because $x e^{-x^2}$ decreases much faster than $x \log x$ – reuns Sep 01 '17 at 01:09
  • @Jeremy It is not true that $H(+\infty x) = 2\pi$ for $x < 0$ – reuns Sep 01 '17 at 01:12
  • In your example your u-v substitution method doesn't get you very far.

    So you drop the lower half of the bound because for values of $x < 0$ the integral has no value.

    The last integral does not include the upper bound of $\infty$ $$ -\int_0^\infty C \phi'(x) dx= - 2\pi (\phi(\infty) - \phi(0)) $$

     – Jeremy Sep 01 '17 at 01:30
  • @Jeremy You don't get it. $\phi' \in L^1$ implies that $-\int_0^\infty \phi'(x)dx = \phi(0)-\phi(\infty)$ where $\phi(\infty)= \lim_{x \to \infty} \phi(x)$ converges, and $\phi\in L^1$ implies that $\phi(\infty) =0$. – reuns Sep 01 '17 at 01:33
  • @Jeremy I made it clear $H(+\infty x) = \lim_{a \to \infty} H(ax) = 2 \pi$ for $x > 0$ and $0$ for $x < 0$... – reuns Sep 01 '17 at 01:36
  • Ok. I can understand that if an integral has a defined value then it's limits approaching infinity must converge to zero.

    That still does not say that $\phi(x)$ at $x=0$ is $\infty$

     – Jeremy Sep 01 '17 at 01:52
  • @Jeremy You meant why $\phi(\infty) = 0$ ? Because otherwise $\phi$ isn't $L^1$. Read my answer and comments again, I explained everything. – reuns Sep 01 '17 at 02:13
  • The definition of the Dirac is that it is $\infty$ at 0 and 0 everywhere else. It also has unit integration. – Jeremy Sep 01 '17 at 02:15
  • @Jeremy Not really. The definition of the Dirac delta is $\int_{-\infty}^\infty \delta(x) \phi(x)dx = \phi(0)$ whenever $\phi$ is continuous (and compactly supported). – reuns Sep 01 '17 at 02:19
  • @Jeremy Is it clear now ? – reuns Sep 01 '17 at 19:51
  • Thank you. Yes. – Jeremy Sep 05 '17 at 00:14
  • I posted another question about the laws of distribution and integrating in lebesgue spaces.

    https://math.stackexchange.com/questions/2419391/lebesgue-spaces-and-integration-by-parts

    I cannot prove the following statement.

    $\phi' \in L^1$ implies $-\int_0^\infty \phi'(x)dx = \phi(0)-\phi(\infty)$

     – Jeremy Sep 12 '17 at 18:45
  • @Jeremy I already answered to that. $\phi' \in L^1$ implies $ \int_{-\infty}^\infty \phi'(x)dx= \lim_{A \to \infty}\int_{-A}^A \phi'(x)dx= \lim_{A \to \infty} \phi(A)-\phi(-A)$ converges. If the limit is not zero then $\phi$ can't be $L^1$. – reuns Sep 12 '17 at 23:02
  • Why is it true that if the limit is non-zero then $\phi \notin L^1$ space? – Jeremy Sep 13 '17 at 00:40
  • What unique property about $\phi$ causes its unbounded limit to approach zero? – Jeremy Sep 13 '17 at 00:41
  • @Jeremy : Come on ! If the limit converges $\lim_{x \to \infty} \phi(x) = C$ with $C\ne 0$ then $\int_{-\infty}^\infty |\phi(x)| dx = \infty$ ie. $\phi \not \in L^1$ – reuns Sep 13 '17 at 00:43
  • The contrapositive of that statement describes why $\phi \in L^1$

    If $\int_{-\infty}^\infty | \phi(x) | dx \ne \infty$ then $\lim_{x \to \infty} \phi(x) = 0$

     – Jeremy Sep 13 '17 at 00:51
  • @Jeremy What are you talking about ? Do you understand why $\int_a^\infty |C| dx = \infty$ if $C \ne 0$ ? – reuns Sep 13 '17 at 00:55
  • Yes. Thank you. – Jeremy Sep 13 '17 at 01:29
  • Let us continue this discussion in chat. – Jeremy Sep 13 '17 at 01:29
  • After looking this over again, I have a question. – Jeremy Jun 16 '20 at 00:13
  • How can the following equation be true? $$ -\int_{-\inf}^{\inf} H(+\inf x)\phi'(x)dx =-\int_{0}^{\inf} 2\pi\phi'(x)dx$$. $2\pi$ is equal to the integral of the sinc. Substituting $2\pi$ for a function inside the integrand with respect to x will change the entire integral. – Jeremy Jun 16 '20 at 00:18
  • $\phi$ is a function of $x$. – Jeremy Jun 16 '20 at 00:24
  • @reuns, There are so many comments here. This leads me to believe that the solution is not as strong as it could be. May I suggest that you add additional comments to your answer. Would you consider re-writing your solution in a more thorough manner? – Michael Levy Apr 16 '21 at 12:45
  • @MichaelLevy No my answer is perfectly fine, the OP was stuck at some very basic facts. – reuns Apr 16 '21 at 12:47
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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 $\epsilon \to 0$: \begin{align*} \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-ix/\epsilon)^2} e^{-\frac12 x^2/\epsilon} \, dk \\ \\ &= \lim_{\epsilon\to 0}e^{-\frac12 x^2/\epsilon} \int_{-\infty}^{\infty} e^{-\frac12\epsilon (k-ix/\epsilon)^2} \, dk \\ &= \lim_{\epsilon\to 0}\sqrt{\frac{2\pi}{\epsilon}} \, e^{-\frac12 x^2/\epsilon} \\ &= 2\pi \, \delta(x) \end{align*}

Michael Levy
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md2perpe
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  • 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

    $$-\frac12\epsilon k^2 + ikx = -\frac12 \epsilon(k^2 - \frac{ix}\epsilon)^2 -\frac12 \frac{x^2}\epsilon$$

     – Jeremy Aug 29 '17 at 03:39
  • Can you please explain the first equality? How did you move from eq. 1 to eq. 2? – Jeremy Aug 29 '17 at 03:54
  • 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, $$\frac12 \epsilon \left( \frac{ix}{\epsilon} \right)^2 = - \frac12 \epsilon \frac{x^2}{\epsilon^2} = -\frac{x^2}{2\epsilon}$$

     – md2perpe Aug 29 '17 at 05:10
  • @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) $\int_{-\infty}^\infty e^{ikx}dk = \lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\frac12\epsilon k^2} e^{ikx} , dk$ – reuns Aug 29 '17 at 11:12
  • 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 $R \to \infty$. – md2perpe Aug 29 '17 at 20:52
  • And for the limit, first of all, the left hand side doesn't have a meaning as an ordinary integral, while the right hand side at least has some meaning as the integral is defined for every $\epsilon>0$ and $x \in \mathbb R$. It can be compared to when the Fourier transform is defined for $L^2$ functions in the same way, i.e. by introducing a damping factor $\exp(-\epsilon x^2)$. – md2perpe Aug 29 '17 at 21:05
  • In you original post you have a typo. $(k^2 - \frac{ix}{\epsilon})^2$ is $(k - \frac{ix}{\epsilon})^2$ – Jeremy Aug 31 '17 at 21:31
  • Thanks, @Jeremy. Actually I had noticed it, but obviously I forgot to fix it. Now I did. – md2perpe Aug 31 '17 at 21:49
  • The integral in your original post the integral converges to $\sqrt{\frac{2\pi}{\epsilon}}$.

    How do you find the integral $$\int_{-\infty}^\infty e^{-\frac12\epsilon(k-\frac{ix}\epsilon)^2} , dk$$

     – Jeremy Aug 31 '17 at 23:39
  • $$\int_{-\infty}^\infty e^{-\frac12\epsilon(k-\frac{ix}\epsilon)^2} , dk = \int_{-\infty}^\infty e^{-\frac12\epsilon k^2} , dk = { t = \sqrt{\epsilon/2} k } = \int_{-\infty}^\infty e^{-t^2} , dt/(\sqrt{\epsilon/2}) = \sqrt{2\pi/\epsilon} $$ If $ix/\epsilon$ had been a real number the first step had been easily justified. Now it's however imaginary which means that the translation occurs in the complex plane and we need to justify it using a complex contour integral. But that's also easily done. – md2perpe Sep 01 '17 at 05:09
  • Why is the imaginary component of the integral not included in the intermediate integral? – Jeremy Sep 05 '17 at 00:22
  • @Jeremy. The first step is just removing the translation of the integrand: $$\int_{-\infty}^\infty e^{-\frac12\epsilon(k-\frac{ix}\epsilon)^2} , dk = { u = k-\frac{ix}\epsilon } = \int_{-\infty}^\infty e^{-\frac12\epsilon u^2} , du$$ Here I have introduced a new variable, but above I kept the same name. – md2perpe Sep 05 '17 at 05:07
  • @md2perpe Can you justify getting from the penulitmate step to the last step. how should we know that its $2,\pi,\delta(x)$, rather than say $k,\delta(x)$? – Michael Levy Apr 16 '21 at 21:14
  • @MichaelLevy. What is the value of the integral of $\sqrt{\frac{2\pi}{\epsilon}} , e^{-\frac12 x^2/\epsilon}$? – md2perpe Apr 16 '21 at 21:31
  • @md2perpe : I see what you indicating. In my opinion, your answer would be improved if you explicitly indicated this and explained your reasoning. – Michael Levy Apr 18 '21 at 01:45