Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A question about integral.

  1. Oct 18, 2013 #1
    Please demonstrate this expression for me:
    For any smooth function f(τ):
    f(+[itex]\infty[/itex])+f(-[itex]\infty[/itex])=lim[itex]_{\epsilon\rightarrow0+}[/itex][itex]\epsilon[/itex][itex]\int[/itex][itex]^{+\infty}_{-\infty}[/itex]dτf(τ)exp(-ε/τ/).
     
  2. jcsd
  3. Oct 19, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    You mean:
    $$\lim_{x\rightarrow\infty}\big(f(x)+f(-x)\big)
    = \lim_{\epsilon\rightarrow 0^+}\epsilon \int_{-\infty}^\infty f(\tau)e^{-\epsilon/\tau}d\tau$$
    ... 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

    vanhees71

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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").
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook