Proving $f(+\infty)+f(-\infty )=\lim_{\varepsilon\to0^+} \varepsilon\int_{-\infty }^\infty {d\tau}f(\tau){e^{-\varepsilon|\tau|}}$ In the 9th chapter of the Weinberg's QFT book, I encounter a formula

$$f(+\infty)+f(-\infty)=\lim_{\varepsilon\to 0^+}\varepsilon\int_{-\infty}^\infty{d\tau}f(\tau){e^{-\varepsilon|\tau|}}
$$
  for any smooth function $f(\tau)$.

If $f(\tau)$ can be expanded by
$$f(\tau) = \sum_{n=0}^\infty \frac{a_n}{n!} \tau^n,$$
I can give the right hand side (before taking $\epsilon \to 0+$) by
$$\sum_{n=0}^\infty \frac{a_n}{\epsilon^n},$$
but I still cannot reach this formula.
Could anyone give me proof of this? Thanks!
 A: Note that $f$ does not need to be smooth for this equation to hold. The continuity of $f$ and the existence of the limits $f(\pm \infty)$ is sufficient. In particular, this condition implies that $f$ is bounded, i.e. $\lVert f \rVert_\infty < \infty$ .
For $t \in \mathbb{R}$ and $\varepsilon > 0$ we have
$$ \tag{1} \left\lvert f \left(\frac{t}{\varepsilon}\right) \mathrm{e}^{- \lvert t \rvert}\right\rvert \leq \lVert f \rVert_\infty \mathrm{e}^{- \lvert t \rvert} $$
and
$$ \tag{2}\lim_{\varepsilon \to 0^+} f \left(\frac{t}{\varepsilon}\right) \mathrm{e}^{- \lvert t \rvert} = \begin{cases} f(\infty) \mathrm{e}^{-t} &, t > 0 \\ f(0) &, t = 0 \\ f(-\infty) \mathrm{e}^t &, t < 0 \end{cases} \, .$$
The function on the right-hand side of $(1)$ is integrable over $\mathbb{R}$, so we can use the dominated convergence theorem to interchange limit and integration in the following calculation:
\begin{align}
\lim_{\varepsilon \to 0^+} \varepsilon \int \limits_{-\infty}^\infty f (\tau) \mathrm{e}^{- \varepsilon \lvert \tau \rvert} \, \mathrm{d} \tau &\stackrel{\varepsilon \tau = t}{=} \lim_{\varepsilon \to 0^+} \int \limits_{-\infty}^\infty f \left(\frac{t}{\varepsilon}\right)\mathrm{e}^{- \lvert t \rvert} \, \mathrm{d} t \stackrel{(1), \text{DCT}}{=} \int \limits_{-\infty}^\infty \lim_{\varepsilon \to 0^+}  f \left(\frac{t}{\varepsilon}\right)\mathrm{e}^{- \lvert t \rvert} \, \mathrm{d} t \\
&\,\,\stackrel{(2)}{=} \int \limits_0^\infty f(\infty) \mathrm{e}^{-t} \, \mathrm{d} t + \int \limits_{-\infty}^0 f(-\infty) \mathrm{e}^t \, \mathrm{d} t = f(\infty) + f(-\infty) \, .
\end{align}
