Contour integral and delta function

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  • #1
StatusX
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I have two related questions. First of all, we have the identity:

[tex] \int_{-\infty}^{\infty} e^{ikx} dk = 2 \pi \delta(x) [/tex]

I'm wondering if it's possible to get this by contour integration. It's not hard to show that the function is zero for x non-zero, but the behavior at x=0 is bugging me.

Next, you can similarly derive something like:

[tex] \int_{0}^{\infty} e^{ikx} dk = \frac{i}{x} + \pi \delta(x) [/tex]

Now if I integrate both sides with respect to z, I get:

[tex] \int_{0}^{\infty} e^{ikx} \frac{dk}{k} = - \ln(x) + i \pi \theta(x) [/tex]

But the integral on the LHS can be rewritten with a change of variables so that it doesn't depend on x. How can this be?
 

Answers and Replies

  • #2
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Contour Integration? Sounds interesting, but I can't guess which representation of the delta function would make the connection. Maybe you will have some luck looking for a proof involving this related fact:

http://mathworld.wolfram.com/FourierTransformDeltaFunction.html

Mathworld gives the citation

Bracewell, R. "The Impulse Symbol." Ch. 5 in The Fourier Transform and Its Applications, 3rd ed. New York: McGraw-Hill, pp. 69-97, 1999.

which sounds like an interesting read.
 
  • #3
StatusX
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Anyone?
 
  • #4
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Ok I apologize for not giving a very well-thought-out answer but I'm in a bit of a hurry. I'll come back later and try to give you some more detail.

So, I am not entirely sure what you're trying to do. I personally feel it is a mistake to treat the delta function like a regular function and to write non-convergent integrals and treat them the same way you would convergent ones. The delta function is actually a distribution (it is a tempered distribution) and I find that the best way to view distributions is as linear functionals on a space of test functions. (You can also think of them as "limits" of sequences of functions which are bounded in L^2 but not L^\infty. This may be somewhat more intuitive.)

Anyway the advantage to either of these views is that it allows you to handle distributions the way you would any other function (by messing with the test function/with the functions in the sequence rather than the distribution directly). Then you should be able to see why these sorts of results hold. It may also help you figure out what's going on in your last line.

Regarding contour integration: probably not helpful, except to the extent that you kinda use that to prove stuff about the Fourier transform...

As I said I'll come back and see whats actually going on later after lunch and a meeting.
 
  • #5
Chris Hillman
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Well, not entirely like other functions; multiplying delta functions is a well-known recipe for disaster, for example (although multiply Heaviside functions is just fine).
 
  • #6
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Now if I integrate both sides with respect to z, I get:

[tex] \int_{0}^{\infty} e^{ikx} \frac{dk}{k} = - \ln(x) + i \pi \theta(x) [/tex]

But the integral on the LHS can be rewritten with a change of variables so that it doesn't depend on x. How can this be?
OK it looks like you've exchanged the order of integration, but your integral doesn't actually converge. That's why you have a problem. See "[URL [Broken] article[/URL] on Fubini's theorem.
 
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