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Annoying Integral

  1. Dec 13, 2006 #1
    I realize this is technically strictly a Math problem, but as it comes up in QFT I was thinking someone here might be more familiar with the theorem. I am trying to prove:

    [tex]\lim_{x \to \infty}f(x) + \lim_{x \to -\infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}[/tex]

    It crops up in Weinberg's QFT book (Vol. I) as he explains how to derive the S matrix using path integrals. f(x) is a wavefunction and for the sake of argument I am making two assumptions:
    1) [tex]\lim_{x \to \pm \infty}f(x)[/tex] both exist and
    2) f(x) is a [tex]C^{\infty}[/tex] function.
    (The text actually only requires f(x) to be "sufficiently smooth." I'll happily accept a proof with a [tex]C^{\infty}[/tex] function and worry about other details later.)

    The specific text reference is "The Quantum Theory of Fields," Vol 1., by Weinberg. It is equation 9.2.15 on page 388.

    Thanks and Merry Christmas everyone!

    Edit: I find it amusing that, though the Math problem is possibly at undergraduate level that someone thought a Physics reference to Quantum Field Theory would be at an undergraduate level and thus moved this post here.
    Last edited: Dec 13, 2006
  2. jcsd
  3. Dec 13, 2006 #2
    The first thing that you might try doing is something that seems rather obvious: split [tex]\int_{-\infty}^\infty \rightarrow \int_{-\infty}^0 + \int_0^{\infty}[/tex] and see if that gets you anywhere. At least that gets rid of the absolute value in the exponent, and I always hate those. My guess is that you might then be able to recognize the exponentials times [tex]\epsilon[/tex] terms as representations of delta functions.
    Last edited: Dec 13, 2006
  4. Dec 13, 2006 #3
    I can't see how it could be related to a delta function since the argument of the exponential is real. However, I've tried two approaches along those lines. The first was to try to work on it as a Laplace transform, which eventually gives me (not including the [tex]\epsilon[/tex] limit)
    [tex]\int_{-\infty}^{\infty}dx f(x) e^{-\epsilon |x|} = \int_0^{\infty} dx f(-x) e^{-\epsilon x} + \int_0^{\infty} dx f(x) e^{-\epsilon x}[/tex]
    which I can't figure out what to do with because I don't know any relationship between the first integral and [tex]L \{ f(x) \} [/tex].

    The second approach was simply to integrate by parts:
    [tex]\int_{-\infty}^{\infty}dx f(x) e^{-\epsilon |x|} = \lim_{x \to \infty} \left [ (f(x) + f(-x)) \frac{e^{-\epsilon x}}{\epsilon} \right ] -[/tex] [tex] \int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}[/tex]

    Applying the [tex]\lim_{\epsilon \to 0^+} \epsilon[/tex] operator to this gives:
    [tex]\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty}dx f(x) e^{-\epsilon |x|} = - \lim_{\epsilon \to 0^+} \lim_{x \to \infty} \left [ (f(x) + f(-x)) e^{-\epsilon x} \right ] - [/tex] [tex]\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}[/tex]

    I am presuming the limit of the integral goes to 0 (I'll work on that later), but I don't know how to take that combined limit of the first term.

    Last edited: Dec 13, 2006
  5. Dec 13, 2006 #4


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    As x->infinity, [itex]e^{-\epsilon x}[/itex]->1, all epsilon>0. Also, be careful with the cusp at x=0 when integrating by parts.
  6. Dec 14, 2006 #5
    Actually that would be "as x -> infinity [itex]e^{-\epsilon x}[/itex] -> 0." (Which is disappointing as the form otherwise was close to what I needed to show.)

    Thanks for the tip about the cusp at x = 0. I always forget about that detail.

  7. Dec 14, 2006 #6
    I just looked at this briefly. If you split the two integrals, as suggested above, and then do a variable change in the two integrals [tex] \bar x \rightarrow \epsilon x [/tex].
    Then we get
    [tex] \lim_{\epsilon \rightarrow 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|} = \lim_{\epsilon \rightarrow 0^+} \epsilon \left( \int_{-\infty}^0 dx f(x) e^{\epsilon x} + \int_0^\infty dx f(x) e^{-\epsilon x} \right) = \lim_{\epsilon \rightarrow 0^+} \left( \int_{-\infty}^0 d\bar x f(\frac{\bar x}{\epsilon}) e^{\bar x} + \int_0^\infty d\bar x f(\frac{\bar x}{\epsilon}) e^{-\bar x}\right).[/tex]

    Now, in a typical physical way to treat mathematics, we can move in the limit inside the two integrals. The limiting procedure then becomes independent of the integration variable. For example the first integral is
    [tex] \int d\bar x \left( \lim_{\epsilon \rightarrow 0^+} f(\frac{\bar x}{\epsilon}) \right) e^{\bar x} = \left( \lim_{x\rightarrow -\infty} f(x) \right) \int_{-\infty}^0 d\bar x e^{\bar x} = \lim_{x \rightarrow -\infty} f(x) [/tex],
    where it is used that the integral over the exponential is 1. The second integral follows in a similar manner.

    This, of course, is not a rigorous mathematical proof, but it is a motivation for why one should believe in the result.
    Last edited: Dec 14, 2006
  8. Dec 14, 2006 #7


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    Oh, right, sorry. Even better though. Then the first term drops. I think you're missing a 1/epsilon on the second term. Then if you can justify switching the limit and integral, which should be possible using the dominated convergence theorem, you just get the infinite integral of f'(x), which will give you you're result up to a wrong sign. I think this will be fixed by more careful treatment of the cusp.
  9. Dec 14, 2006 #8
    Hmm.. I'm not sure where you mean there should be a [tex] 1 / \epsilon [/tex]. I think that what I wrote is correct.
    I might have been a little unclear. The variable change should be [tex] \bar x = \epsilon x [/tex], which gives us [tex] d\bar x = \epsilon dx [/tex].
    The different signs comes from the fact that the integrals are over the negative and positive parts of the real axis, respectively.
  10. Dec 14, 2006 #9


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    Sorry, I was referring to the last line in post 3.
  11. Dec 29, 2006 #10
    Sorry it took me so long to answer, I was away over Christmas.

    Not rigorous, but quite good enough for my puposes. Thanks! (And thanks also to StatusX for your efforts as well!)

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