Dirac Delta and Residue Calculus

[Daniel Gallimore]

I'm an undergraduate student, so I understand that it may be difficult to provide an answer that I can understand, but I have experience using both the Dirac delta function and residue calculus in a classroom setting, so I'm at least familiar with how they're applied.

Whether you're integrating along a closed loop around a singularity in the complex plane or you're integrating on a closed sphere in 3D space about a Dirac delta (like you might do in E&M), the value of the integral depends entirely on the point where the singularity/delta is located. Do these similarities betray a connection between the Dirac delta and residue calculus?

 

[stevendaryl]

I'm an undergraduate student, so I understand that it may be difficult to provide an answer that I can understand, but I have experience using both the Dirac delta function and residue calculus in a classroom setting, so I'm at least familiar with how they're applied.

Whether you're integrating along a closed loop around a singularity in the complex plane or you're integrating on a closed sphere in 3D space about a Dirac delta (like you might do in E&M), the value of the integral depends entirely on the point where the singularity/delta is located. Do these similarities betray a connection between the Dirac delta and residue calculus?

Well, I would say that there is a closer connection between residues and the Heaviside function [itex]H(x)[/itex] defined as follows:

[itex]H(x) = 0[/itex] if [itex]x < 0[/itex]
[itex]H(x) = 1[/itex] if [itex]x > 0[/itex]

Then an integral representation of [itex]H(x)[/itex] is:

[itex]H(x) = lim_{\epsilon \rightarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{+\infty} \frac{1}{\tau -i \epsilon} e^{ix\tau} d\tau[/itex]

which you can prove using residues.

[itex]H(x)[/itex] is related to the delta-function by formally taking the derivative of this integral representation with respect to [itex]x[/itex]:

[itex]\delta(x) \equiv \frac{dH}{dx} = lim_{\epsilon \rightarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{+\infty} \frac{i \tau}{\tau -i \epsilon} e^{ix\tau} d\tau[/itex]
[itex]= \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ix\tau} d\tau[/itex]

 

[zwierz]

It is hard to encounter adequate understanding of the distributions concept in physical textbooks.
Try
KöSaku Yosida. Functional Analysis. Sixth Edition. Springer-Verlag. Berlin Heidelberg New York 1980

 

[zwierz]

is related to the delta-function by formally taking the derivative of this integral representation with respect to xx:

δ(x)≡dHdx=limϵ→012πi∫+∞−∞iττ−iϵeixτdτ\delta(x) \equiv \frac{dH}{dx} = lim_{\epsilon \rightarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{+\infty} \frac{i \tau}{\tau -i \epsilon} e^{ix\tau} d\tau
=12π∫+∞−∞eixτdτ

it just remains to explain what this divergent integral means and in which sense the limit ##\epsilon\to 0## is understood