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An identity involving a Dirac delta function.

  1. Jul 29, 2009 #1
    I have been reading papers for my research and I came across this equation twice:

    [tex]\lim_{\eta\to 0+}\frac{1}{x+i \eta} = P\left(\frac{1}{x}\right) - i \pi \delta(x)[/tex]

    Where P is the pricipal part.

    It has been quite a while since I have had complex variables, but might it come from the residue theorem? If anyone knows the origin of this identity and how it is derived, I would love to see it.

    Thanks!


    Edit: added images

    Here are clips from the two papers below. In the second clip, the sum over eta is implied.
     

    Attached Files:

    Last edited: Jul 29, 2009
  2. jcsd
  3. Jul 29, 2009 #2
    How odd. When x is nonzero, the LHS reduces to 1/x. The only case where the limit on the LHS has any significance is when x=0.... but in that case, the RHS contains a division by zero!

    Either there's some missing context, or someone is producing useless equations for the soul purpose of confusing you ;-)
     
  4. Jul 29, 2009 #3
    Looks very interesting equation, although I have difficulty understanding what it means. Did I understand correctly, that it means this:

    [tex]
    \lim_{\eta\to 0^+} \int\limits_{[-r,r]} \frac{f(x)}{x + i\eta} dx \quad=\quad \lim_{\epsilon\to 0^+} \int\limits_{[-r,-\epsilon]\cup [\epsilon,r]} \frac{f(x)}{x} dx \quad-\quad i\pi f(0)
    [/tex]

    ?
     
  5. Jul 29, 2009 #4
    For sake of testing I substituted a test function [itex]f(x)=f(0) + f'(0)x[/itex]. Direct integrations give

    [tex]
    \int\limits_{[-r,r]} \frac{f(0)+f'(0)x}{x+i\eta}dx = f(0)\big(\log(r+i\eta) - \log(-r+i\eta)\big) \;+\; f'(0)\big(2r - i\eta\log(r+i\eta) + i\eta\log(-r+i\eta)\big) \underset{\eta\to 0+}{\to} -i\pi f(0) + 2r f'(0)
    [/tex]

    [tex]
    \int\limits_{[-r,-\epsilon]\cup [\epsilon,r]} \frac{f(0) + f'(0)x}{x} = f(0)\big(\log |\epsilon| - \log |r|\big) \;+\; f'(0)(r-\epsilon) \;+\; f(0)\big(\log(r) - \log(\epsilon)\big) \;+\; f'(0)(r-\epsilon) = f'(0)(2r-2\epsilon) \underset{\epsilon\to 0^+}{\to} 2r f'(0)
    [/tex]

    So that's the way it seems to be.
     
  6. Jul 30, 2009 #5

    Mute

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    Homework Helper

    The identity exists only under an integral sign. If you were to perform the integral

    [tex]\int_{-\infty}^{\infty} dx~\frac{f(x)}{x + i\eta}[/tex]

    over the whole real line as [itex]\eta \rightarrow 0^+[/itex] using contour integration with a contour with a small circular arc around the singular point, you would find the result to be

    [tex]\mathcal{P}\int_{-\infty}^{\infty} dx~\frac{f(x)}{x} -i\pi f(0)[/tex]

    where the "P" denotes the principal part integral. This entire result can be put under an integral sign using the delta function:

    [tex]\int_{-\infty}^{\infty} dx~f(x)\left[\mathcal{P}\left(\frac{1}{x}\right) -i\pi \delta(x) \right][/tex]

    Hence dropping the integral sign gives the identity, but the identity is only valid when integrating. It has no meaning otherwise.
     
    Last edited: Jul 30, 2009
  7. Jul 30, 2009 #6
    Anyone knowing how to prove that?
     
  8. Jul 30, 2009 #7

    Mute

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    By contour integration, just like I said.

    Consider the integral

    [tex]\int_C dz~\frac{f(z)}{z + i\eta},[/tex]

    where C is the contour composed of a straight line paralell to the axis at [itex]-i\eta[/itex] from -R to -epsilon, a circular arc of radius epsilon going over the pole at [itex]z=-i\eta[/itex], another straight line from epsilon to R, then a large circular arc of radius arc to close the contour.

    The contour encloses no poles, so the contour integral is zero (though that doesn't actually matter here). Now, decompose the countour integral into integrals along each segment:

    [tex]\int_C dz~\frac{f(z)}{z + i\eta} = \int_{-R}^{-\epsilon}dx~\frac{f(x)}{x + i\eta}+ \int_{\epsilon}^{R}dx~\frac{f(x)}{x + i\eta} + i \int_{\pi}^{0}d\theta~\epsilon e^{i\theta}\frac{f(\epsilon e^{i\theta})}{\epsilon e^{i\theta} + i\eta} + i \int_{0}^{\pi}d\theta~R e^{i\theta}\frac{f(R e^{i\theta})}{R e^{i\theta} + i\eta}[/tex]

    Under the assumption that [itex]f(Re^{i\theta}) \rightarrow 0[/itex] as [itex]R \rightarrow \infty[/itex], the integral along the large arc vanishes. As eta goes to zero the integral along the small arc gives [itex]-i\pi f(0)[/itex], and the remaing two terms give the principal value term as R goes to infinity and epsilon goes to zero. After some cosmetics wherein the f(0) is put back into the integral using the delta function, the identity is obtained.
     
  9. Jul 30, 2009 #8

    Hans de Vries

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    Science Advisor

    Hi

    The eta here is just like the epsilon used in the (causal) propagators
    in QFT. The two terms on the RHS can be seen as a Hilbert pair related
    by the Hilbert transform. The latter is for me personally the preferred way
    to handle the propagators. See for instance section 1.15 discussing both
    for the causal photon propagator in chapter 1 of my book here:

    http://physics-quest.org/Book_Chapter_EM_basic.pdf

    Regards, Hans
     
  10. Jul 31, 2009 #9
    (update begins): Holy smoke! I had forgotten what the notation [itex]f\circ g[/itex] means when functions are composed. Well when two paths [itex]\gamma_1:[0,1]\to\mathbb{C}[/itex] and [itex]\gamma_2:[0,1]\to\mathbb{C}[/itex] are combined to give a mapping

    [tex]
    [0,1]\to\mathbb{C},\quad t\mapsto \left\{\begin{array}{ll}
    \gamma_1(2t),\quad &0\leq t\leq\frac{1}{2}\\
    \gamma_2(2t-1),\quad &\frac{1}{2}\leq t\leq 1\\
    \end{array}\right.
    [/tex]

    I will denote this with [itex]\gamma_2\cup\gamma_1:[0,1]\to\mathbb{C}[/itex], because I don't remember what the correct notation was, if it exists.
    (update ends)

    Mute, that was a confused post. There is two problems with it. (1): We cannot assume that [itex]f[/itex] vanishes sufficiently quickly in directions [itex]re^{i\theta}[/itex], [itex]r\to\infty[/itex], [itex]-\pi\leq \theta\leq 0[/itex]. It is common for an analytic continuation of some test function [itex]f:\mathbb{R}\to\mathbb{R}[/itex] to diverge somewhere at infinity. (2): We are not interested in integrating along a contour that encloses the lower half plane.

    It seems likely, that when the theorem is proven, it will be proven like this: To examine the integral

    [tex]
    \lim_{\eta\to 0^+} \int\limits_{[a,b]}\frac{f(z)}{z+i\eta}dz
    [/tex]

    where [itex]a<0<b[/itex], we choose three pieces of paths, [itex]\gamma_{1,\eta}, \gamma_{2,\eta}, \gamma_{3,\eta}[/itex], smoothly depending on [itex]\eta[/itex], such that they can always be composed as [itex]\gamma_{3,\eta}\cup\gamma_{2,\eta}\cup\gamma_{1,\eta}[/itex] for all [itex]\eta[/itex], such that [itex]\gamma_{1,\eta}[/itex] begins at [itex]a[/itex] for all [itex]\eta[/itex], and such that [itex]\gamma_{3,\eta}[/itex] ends at [itex]b[/itex] for all [itex]\eta[/itex]. Then the previous expression is the same as this

    [tex]
    \lim_{\eta\to 0^+} \int\limits_{\gamma_{3,\eta}\cup\gamma_{2,\eta}\cup\gamma_{1,\eta}} \frac{f(z)}{z+i\eta}dz
    [/tex]

    If the gammas are chosen correctly, then it should be possible to show that

    [tex]
    \lim_{\eta\to 0^+} \Big(\int\limits_{\gamma_{1,\eta}} \frac{f(z)}{z+i\eta}dz + \int\limits_{\gamma_{3,\eta}} \frac{f(z)}{z + i\eta}dz\Big) = \lim_{\epsilon\to 0^+} \int\limits_{[a,-\epsilon]\cup [\epsilon,b]} \frac{f(z)}{z}dz
    [/tex]

    and

    [tex]
    \lim_{\eta\to 0^+}\int\limits_{\gamma_{2,\eta}} \frac{f(z)}{z + i\eta}dz = -i\pi f(0).
    [/tex]
     
    Last edited: Jul 31, 2009
  11. Jul 31, 2009 #10

    Mute

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    Homework Helper

    You're right, that was a confused post. Forget the part about the large circular arc (which, by the way, enclosed the upper half plane, not the lower, though that's irrelevant now), and consider the remaining contour I specified: a piece from a to -epsilon, then the small circular arc of radius epsilon going over the pole, and then another straight arc from epsilon to b. Take the limit as eta, then epsilon go to zero and it should give the identity, as can be seen in my post above by ignoring the large arc term and treating C as this contour.

    Hence, now that should be an acceptable proof that actually proves what we want.

    Coincidentally, I only now just managed to find the wikipedia article on this, which gives another proof.

    http://en.wikipedia.org/wiki/Sokhatsky–Weierstrass_theorem
     
  12. Jul 31, 2009 #11
    It's an idea for a proof, but there are still problems to be dealt with before it's a real proof. It is difficult stuff when the integration contour and the integrand both change as a function of some parameter (now [itex]\eta[/itex]) simultaneously.

    That looks good. In what I explained, I was still assuming that [itex]f:\mathbb{R}\to\mathbb{R}[/itex] could be analytically continued to some small environment of the real axis. The approach in the Wikipedia page does not need this assumption. The theorem clearly will not really need analyticity of [itex]f[/itex].

    The Wikipedia page however does not give satisfying explanation for the result

    [tex]
    \lim_{\epsilon\to 0^+} \int\limits_a^b \frac{x^2}{x^2+\epsilon^2} \frac{f(x)}{x}dx = \lim_{\epsilon\to 0^+} \int\limits_{[a,-\epsilon]\cup [\epsilon,b]} \frac{f(x)}{x}dx
    [/tex]
     
  13. Jul 31, 2009 #12

    Mute

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    Well, sure, if you want absolute rigor. ;) Since from a physicist's perspective that would likely be a sufficient proof, I left it at that. I did suspect that a mathematician might be unsatisfied with the proof. However, since such changing contours are common in situations such as pacman contours about a branch cut, and we know those are generally fine and well behaved, I didn't bother to think much about a rigorous proof. (And I am not sure I could have devised one).

    If you want a rigorous proof, this result follows from the Plemelj (or Sokhatsky-Plemelj?) formulas, but I don't know where you might find a proof for those off-hand.

    The Plemelj formula are: For a function

    [tex]f(\zeta) = \frac{1}{\pi}\int_\Gamma dz~\frac{\rho(z)}{z-\zeta},[/tex]

    with [itex]\Gamma[/itex] a segment of the real axis, the following relations hold:

    [tex]\frac{1}{2}(f(x+i\epsilon) + f(x-i\epsilon)) = i\rho(x),[/tex]
    [tex]\frac{1}{2}(f(x+i\epsilon) - f(x-i\epsilon)) = \frac{1}{\pi}\mathcal{P}\int_{\Gamma}dx'~\frac{\rho(x')}{x'-x},[/tex]

    under "relatively mild conditions" on rho(x), and the limit being taken as epsilon goes to zero.
     
  14. Jul 31, 2009 #13
    I succeeded proving this now. All comments about contour integration should be forgotten. The result is true for all continuous functions, even for those which are not differentiable. So one should not get into complex analysis. The Wikipedia page about Sokhatsky-Weierstrass theorem outlines a correct way, which begins with:

    [tex]
    \frac{1}{x \pm i\epsilon} = \mp \frac{i\epsilon}{x^2 + \epsilon^2} + \frac{x}{x^2 + \epsilon^2}
    [/tex]

    We can expect that the reader knows how to prove

    [tex]
    \frac{\epsilon}{x^2 + \epsilon^2} \underset{\epsilon\to 0^+}{\to} \pi \delta(x).
    [/tex]

    So the real task we are left with is to prove this:

    If we assume it known that the principal value exists, then this equation is equivalent with the equation

    [tex]
    \lim_{\epsilon\to 0^+}\int\limits_a^b\Big( \frac{x^2}{x^2 + \epsilon^2} \;-\; \big(1 - \chi_{[-\epsilon,\epsilon]}(x)\big)\Big)\frac{f(x)}{x} dx =: \lim_{\epsilon\to 0^+} I(\epsilon) = 0
    [/tex]

    For symmetry reasons, it is sufficient to show that this integral vanishes when it is integrated over the values [itex]x>0[/itex]. So we can set [itex]a=0[/itex]. I split the integral in two pieces, over [itex][0,\epsilon][/itex] and over [itex][\epsilon,b][/itex].

    [tex]
    \int\limits_0^{\epsilon} \frac{x}{x^2+\epsilon^2} f(x) dx
    [/tex]

    and

    [tex]
    \int\limits_{\epsilon}^b \Big(\frac{x^2}{x^2 + \epsilon^2} - 1\Big)\frac{f(x)}{x} dx = -\int\limits_{\epsilon}^b \frac{\epsilon^2}{x^2 + \epsilon^2} \frac{f(x)}{x} dx
    [/tex]

    The integral over the [itex][0,\epsilon][/itex] can be calculated like this:

    [tex]
    (\inf_{0<x<\epsilon}f(x)) \int\limits_0^{\epsilon} \frac{x}{x^2 + \epsilon^2} dx \leq \int\limits_0^{\epsilon}\frac{xf(x)}{x^2 + \epsilon^2} dx \leq (\sup_{0<x<\epsilon}f(x)) \int\limits_0^{\epsilon} \frac{x}{x^2 + \epsilon^2} dx
    [/tex]

    The infimum and supremum are going to approach [itex]f(0)[/itex].

    [tex]
    \int\limits_0^{\epsilon} \frac{x}{x^2 + \epsilon^2} dx = \frac{1}{2}\log(2)
    [/tex]

    So the integral over [itex][0,\epsilon][/itex] approaches the number [itex]\frac{1}{2}\log(2)f(0)[/itex]. The integral over [itex][\epsilon, b][/itex] can be calculated with a variable change [itex]x=\epsilon y[/itex]. It becomes

    [tex]
    -\int\limits_1^{b/\epsilon} \frac{f(\epsilon y)}{(1+y^2)y} dy
    [/tex]

    If [itex]f[/itex] is bounded on [itex][a,b][/itex], which it is since it is continuous, then we can take the limit [itex]f(\epsilon y)\to f(0)[/itex] outside the integral. We get

    [tex]
    -f(0) \int\limits_1^{\infty} \frac{1}{(1+y^2)y} dy = -f(0)\int\limits_1^{\infty}\Big(\frac{1}{y} \;-\; \frac{y}{1+y^2}\Big)dy = -\frac{1}{2}\log(2)f(0) \;-\; \lim_{R\to\infty}\Big(\log(R) - \log(\sqrt{1 + R^2})\Big) = -\frac{1}{2}\log(2)f(0)
    [/tex]

    So the integrals over [itex][0,\epsilon][/itex] and [itex][\epsilon,b][/itex] cancel on the limit [itex]\epsilon\to 0^+[/itex], and we get the desired result [itex]I(\epsilon)\to 0[/itex].
     
    Last edited: Jul 31, 2009
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