Identify for $f(\infty)+f(-\infty)$ in quantum field theory In Matthew Schwartz's textbook, Quantum Field Theory and the Standard Model, equation 14.68 on page 266 says the following:
$$f(\infty)+f(-\infty)=\lim_{\varepsilon\rightarrow0^+}\varepsilon\int_{-\infty}^{\infty}dtf(t)e^{-\varepsilon|t|}.\tag{14.68}$$
The only constraint stipulated in the text is that $f(\tau)$ must be a smooth function. Can anyone help me to prove this expression? When I try to evaluate the right side by bringing the limit and the $\varepsilon$ inside the integral, I find that the right side is equal to zero. I've tried using Dirac delta function identities to prove this, but I've had no luck.
 A: $$
\lim_{\varepsilon\rightarrow0^+}\varepsilon\int_{-\infty}^{\infty}dtf(t)e^{-\varepsilon|t|}$$
$$
=\lim_{\varepsilon\rightarrow0^+}\varepsilon\int_{-\infty}^{0}dtf(t)e^{\varepsilon t}+\lim_{\varepsilon\rightarrow0^+}\varepsilon\int_{0}^{\infty}dtf(t)e^{-\varepsilon t}
$$
$$
=\lim_{\varepsilon\rightarrow0^+}\int_{-\infty}^{0}dtf(t)\frac{de^{\varepsilon t}}{dt}-\lim_{\varepsilon\rightarrow0^+}\int_{0}^{\infty}dtf(t)\frac{de^{-\varepsilon t}}{dt}
$$
$$
=-\lim_{\varepsilon\rightarrow0^+}\int_{-\infty}^{0}dt\frac{df(t)}{dt}e^{\varepsilon t}+f(0)+\lim_{\varepsilon\rightarrow0^+}\int_{0}^{\infty}dt\frac{df(t)}{dt}de^{-\varepsilon t}+f(0)
$$
[The f(0) terms above are from the integration by parts, the other boundary terms are zero for finite $\epsilon$... so now we can take $\epsilon$ to zero]
$$
=-\int_{-\infty}^{0}dt\frac{df(t)}{dt}+\int_{0}^{\infty}dt\frac{df(t)}{dt}+2f(0)
$$
$$
=-f(0)+f(-\infty)+f(\infty)-f(0)+2f(0)
$$
$$
=f(\infty)+f(-\infty)
$$
A: Make the change of variable $x=\varepsilon t$, then the integral becomes
$$\lim_{\varepsilon \rightarrow 0^+} \int_{-\infty}^{+\infty}dxf(x/\varepsilon)e^{-|x|}$$
Then (assuming the mathematicians allow it!) take the limit inside the integral to render the $f(x/\varepsilon)$ into $f(\pm\infty)$ (break the integral into [-$\infty$,0] and [0,$\infty$]) and take it out of the integral.
Intuitively, with reference to $f(x/\varepsilon)$, what $\varepsilon$ does is to shrink the function along $x$. Take $\varepsilon$ to zero and what is left is the steady-state values of the function at infinity.
