Integral resulting in delta function.

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Homework Help Overview

The discussion revolves around evaluating an integral involving the Heaviside step function and its relationship to the delta function. The integral in question is presented as an attempt to demonstrate its equivalence to a specific expression involving the delta function and a term dependent on omega.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the transformation of the integral by changing its bounds due to the Heaviside function and discuss the convergence of the integral. There are suggestions to introduce a convergence factor and to consider complex analysis techniques. Questions arise regarding the application of the delta function definitions and the limits involved in the evaluation.

Discussion Status

Some participants have provided guidance on methods to approach the problem, including the use of convergence factors and complex analysis. However, there remains uncertainty about the limits and the rigor of the approaches discussed. Multiple interpretations of the problem are being explored, and while some progress has been made, there is no explicit consensus on the best method to achieve the desired result.

Contextual Notes

Participants note the potential issues with convergence and the behavior of the integrand as parameters approach certain limits. There is also mention of the integral being treated as an operator in omega, which adds complexity to the discussion.

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Homework Statement



Hi All. I am given this integral:

[tex]\int_{-\infty}^{\infty}A\Theta e^{i\omega t}dt[/tex]

I need to show that it's equal to the following:

[tex]=A(\pi \delta(\omega)+\frac{i}{\omega})[/tex]

Homework Equations

Theta is the Heavyside step function.

The Attempt at a Solution



The step function changes the lower bound of the integral to 0:

[tex]\int_{0}^{\infty}Ae^{i\omega t}dt[/tex]

Elementary integration then gives:

[tex]\lim_{t=\infty}\frac{e^{i\omega t}}{i\omega} +\frac{i}{\omega}[/tex]

I'm not sure how to relate that first limit to any of the delta function definitions I know. A hint was given to use the following definition:

[tex]\pi\delta(x)=\lim_{A=0}\frac{A}{x^2+A^2}[/tex]

However, I'm not sure how to apply this limit to help get to the result.

Thanks for any hints/help!
 
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The integral won't converge because the integrand doesn't vanish as t goes to infinity, so throw in a convergence factor e-λt and take the limit as λ→0+:

[tex]\lim_{\lambda \to 0^+} \int_0^\infty Ae^{(i\omega-\lambda)t}\,dt[/tex]

You can transform the result into a form where you'll be able to apply the hint you were given.
 
(Side point: there are convergent integrals whose integrand fails to vanish at infinity)


Since the integral is apparently meant to be an integral of operators in [itex]\omega[/itex] rather than an integral of functions, one approach you might want to take is to show that it operates like the thing you're trying to prove it equal to. That is, compute

[tex]\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} A \Theta(\omega) e^{i \omega t} f(\omega) \, d\omega \, dt[/tex]

I don't see how to use the hint if you tried this approach, though.




Another thing I'd be tempted to try is to invoke complex analysis; integrate around wedge-shaped regions whose corner is at the origin and whose edges are the positive real and imaginary axes, and take the limit as radius goes to infinity. I don't know if distributions and complex residues interact well, though.
 
Vela- Thanks! I was able to get to the result using your method.

Hurkyl- I thought about taking the complex analysis route as well. However, like you said the potential issues with residues and distribution scared me away. I didn't think it was worth the trouble, even though it may end up being more rigorous than vela's approach.

I think vela's approach will suffice for this class, however I am a little worried by some of the limits that appear. For instance, in order to get the correct result, we must have:

[tex]\lim_{\lambda=0^+}\lim_{t=\infty} (i\omega e^{(-\lambda + i\omega )t})=0[/tex]

I am by no means certain that it is the case. i.e The exponential term becomes small as lambda goes to 0, but then grows again as you take t to infinity.
 

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