Can You Show Me How to Find the Integral of a Smooth Function Using Limits?

[ndung200790]

Please demonstrate this expression for me:
For any smooth function f(τ):
f(+\infty)+f(-\infty)=lim_{\epsilon\rightarrow0+}\epsilon\int^{+\infty}_{-\infty}dτf(τ)exp(-ε/τ/).

 

[Simon Bridge]

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 $\pm\infty$.

 

[vanhees71]

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