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/).

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##.

Gold Member
2021 Award
No, what he means is
$$\lim_{\epsilon \rightarrow 0^+}\epsilon \int_{-\infty}^{\infty} \mathrm{d} t \; f(t) \exp(-\epsilon |t|)=f(\infty)+f(-\infty).$$
$$I_1=\epsilon \int_0^{\infty} \mathrm{d} t f(t) \exp(-\epsilon t).$$
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.