S. Weinberg, “The Quantum theory of fields: Foundations” (1995), Eq. (9.2.15) In Weinberg's book The Quantum Theory of Fields, Volume 1 on p.388 (Chapter 9), the following identity is used (with $f$ being any "reasonable" function):
$$f(+\infty) + f(-\infty) = \lim_{\epsilon \rightarrow 0^+} \epsilon \int_{-\infty}^{+\infty} d\tau f(\tau) e^{-\epsilon |\tau|}.\tag{9.2.15} $$
I don't understand the identity in a qualitative / heuristic way.
 A: \begin{equation}
\begin{split}
I &= \lim_{\epsilon\to0} \epsilon \int_{-\infty}^{\infty} \mathrm{d} \tau \, f(\tau) e^{- \epsilon |\tau| } \\
&= \lim_{\epsilon\to0} \epsilon \int_{-\infty}^{0} \mathrm{d}\tau \, f(\tau) e^{\epsilon \tau } + \lim_{\epsilon\to0} \epsilon \int_{0}^{\infty} \mathrm{d}\tau \, f(\tau) e^{- \epsilon \tau } \\
&= \lim_{\epsilon\to0} \int_{-\infty}^{0} \mathrm{d}\tau\, f(\tau) \partial_\tau e^{\epsilon \tau } -\lim_{\epsilon\to0} \int_{0}^{\infty} \mathrm{d}\tau\,  f(\tau) \partial_\tau e^{- \epsilon \tau } \\
&= - \lim_{\epsilon\to0} \int_{-\infty}^{0} \mathrm{d}\tau \, \partial_\tau  f(\tau) e^{\epsilon \tau } + \lim_{\epsilon\to0} \int_{0}^{\infty} \mathrm{d}\tau \, \partial_\tau f(\tau)  e^{- \epsilon \tau }  + 2 f(0) \\
&= -  \int_{-\infty}^{0} \mathrm{d}\tau \, \partial_\tau  f(\tau)   + \int_{0}^{\infty} \mathrm{d}\tau \, \partial_\tau f(\tau) + 2 f(0) \\
&= (- f(0) + f(-\infty) ) + (f(\infty) - f(0) ) + 2 f( 0 ) \\
&= f(\infty) + f(-\infty)
\end{split}
\end{equation}
A: Assume that the limit $f(\infty):= \lim_{t\to\infty}f(t)$ exists.


*

*Method 1:  Assume that $f$ is Lebesgue measurable. Assume that $f\geq 0$ is monotone on $\mathbb{R}_+$ or assume that the net $t \mapsto f(t/\epsilon)$ for $\epsilon>0$ has an integrable majorant on $\mathbb{R}_+$. Then $$
\begin{align}\epsilon\int_{\mathbb{R}}\!dt~f(t)~\theta(t)e^{-\epsilon t} 
~=~&\int_{\mathbb{R}}\!dt~f(t/\epsilon)~\theta(t)e^{-t} \cr
~\longrightarrow~&\int_{\mathbb{R}}\!dt~f(\infty)~\theta(t)e^{-t}\cr 
~=~&f(\infty)\quad\text{for}\quad\epsilon\to 0^+.\tag{1}\end{align}$$

*Method 2:  Assume that $f$ is differentiable and $f^{\prime}$ is integrable on $\mathbb{R}_+$. Then 
$$\begin{align}\epsilon\int_0^{\infty}\!dt~f(t)~e^{-\epsilon t} 
~=~& - \int_0^{\infty}\!dt~f(t)~\frac{d}{dt}e^{-\epsilon t}\cr
~\stackrel{\text{int. by parts}}{=}&~\int_0^{\infty}\!dt~f^{\prime}(t)e^{-\epsilon t}-\left[f(t)~e^{-\epsilon t} \right]_0^{\infty}\cr
~\longrightarrow~&\int_0^{\infty}\!dt~f^{\prime}(t) +f(0) \cr 
~=~&f(\infty)\quad\text{for}\quad\epsilon\to 0^+.\tag{2}\end{align}$$

*Method 3:  (Method 2 rewritten with distributions.) 
$$\begin{align}\epsilon\int_{\mathbb{R}}\!dt~f(t)~\theta(t)e^{-\epsilon t} 
~=~& - \int_{\mathbb{R}}\!dt~f(t)~\theta(t)\frac{d}{dt}e^{-\epsilon t}\cr
~\stackrel{\text{int. by parts}}{=}&~\int_{\mathbb{R}}\!dt~e^{-\epsilon t}\frac{d}{dt}(f(t)~\theta(t))\cr
~\longrightarrow~&\int_{\mathbb{R}}\!dt~\frac{d}{dt}(f(t)~\theta(t))\cr
~=~&\int_{\mathbb{R}}\!dt~\frac{df(t)}{dt}\theta(t)+\int_{\mathbb{R}}\!dt~f(t) \delta(t) \cr
~=~&f(\infty)\quad\text{for}\quad\epsilon\to 0^+.\tag{3}\end{align}$$
A: I can understand the identity qualitatively now:
If the function f(t) has a well defined value at $t = \pm\infty$, then for large values of $|t|$ the function f is essentially constant with values $f(-\infty)$ for $t < 0$ and $f(+\infty)$ for $t > 0$.
For tiny $\epsilon$ the Exponential factor $e^{-\epsilon |t|}$ is essentially equal to $1$ for  $|t| < 1 / \epsilon $, and almost Zero for large $|t| > 1 / \epsilon$.
We then have approximately: $\epsilon\int_{-\infty}^\infty \mathrm{d}t \, f(t) e^{-\epsilon |t|} \approx \epsilon\int_{1 / \epsilon}^0 \mathrm{d}t \, f(-\infty) + \epsilon\int_0^{1 / \epsilon} \mathrm{d}t \, f(+\infty) = f(-\infty) + f(+\infty)$
