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A question about integral.

  1. Oct 18, 2013 #1
    Please demonstrate this expression for me:
    For any smooth function f(τ):
  2. jcsd
  3. Oct 19, 2013 #2

    Simon Bridge

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    You mean:
    = \lim_{\epsilon\rightarrow 0^+}\epsilon \int_{-\infty}^\infty f(\tau)e^{-\epsilon/\tau}d\tau$$
    ... but you can demonstrate it to yourself by picking a function and doing the integration.
    Hint: pick one where you know the value at ##\pm\infty##.
  4. Oct 21, 2013 #3


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    No, what he means is
    [tex]\lim_{\epsilon \rightarrow 0^+}\epsilon \int_{-\infty}^{\infty} \mathrm{d} t \; f(t) \exp(-\epsilon |t|)=f(\infty)+f(-\infty).[/tex]
    We start with one half of the integral
    [tex]I_1=\epsilon \int_0^{\infty} \mathrm{d} t f(t) \exp(-\epsilon t).[/tex]
    Substitution of [itex]t=\epsilon \eta[/itex] leads to
    [tex]I_1=\int_0^{\infty} \mathrm{d} \eta f \left (\frac{\eta}{\epsilon} \right ) \exp(-\eta).[/tex]
    Now according to the mean-value theorem for integration, there exists some [itex]\tilde{\eta} > 0[/itex] such that
    [tex]I_1=f \left (\frac{\tilde{\eta}}{\epsilon} \right ) \int_0^{\infty} \mathrm{d} \eta \exp(-\eta) =f \left (\frac{\tilde{\eta}}{\epsilon} \right ).[/tex]
    Now for [itex]\epsilon \rightarrow 0^+[/itex] this gives [itex]f(\infty)[/itex], supposed this limit exists.

    The other half of the integral can be treated analogously. Such considerations play an important role in scattering theory ("adiabatic switching of the interaction").
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