A question about integral.

  • #1
ndung200790
519
0
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/).
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,874
1,657
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##.
 
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
20,730
11,564
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").
 

Suggested for: A question about integral.

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
936
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
902
Replies
2
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
7
Views
829
Top