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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.

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    $\begingroup$ Please describe what exactly You don't understand and why. It will be easier for people here to help You. $\endgroup$
    – Wojciech
    Mar 27, 2014 at 13:32
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    $\begingroup$ Final value theorem. $\endgroup$
    – Nikolaj-K
    Mar 27, 2014 at 14:26

3 Answers 3

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\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}

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Assume that the limit $f(\infty):= \lim_{t\to\infty}f(t)$ exists.

  1. 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 $$ \begin{align}\epsilon\int_{\mathbb{R}}\!dt~f(t)~\theta(t)e^{-\epsilon t} ~=~&\int_{\mathbb{R}}\!dt~f(t/\epsilon)~\theta(t)e^{-t} \cr ~\longrightarrow~&\int_{\mathbb{R}}\!dt~f(\infty)~\theta(t)e^{-t}\cr ~=~&f(\infty)\quad\text{for}\quad\epsilon\to 0^+.\tag{1}\end{align}$$

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

  3. Method 3: (Method 2 rewritten with distributions.) $$\begin{align}\epsilon\int_{\mathbb{R}}\!dt~f(t)~\theta(t)e^{-\epsilon t} ~=~& - \int_{\mathbb{R}}\!dt~f(t)~\theta(t)\frac{d}{dt}e^{-\epsilon t}\cr ~\stackrel{\text{int. by parts}}{=}&~\int_{\mathbb{R}}\!dt~e^{-\epsilon t}\frac{d}{dt}(f(t)~\theta(t))\cr ~\longrightarrow~&\int_{\mathbb{R}}\!dt~\frac{d}{dt}(f(t)~\theta(t))\cr ~=~&\int_{\mathbb{R}}\!dt~\frac{df(t)}{dt}\theta(t)+\int_{\mathbb{R}}\!dt~f(t) \delta(t) \cr ~=~&f(\infty)\quad\text{for}\quad\epsilon\to 0^+.\tag{3}\end{align}$$

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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: $\epsilon\int_{-\infty}^\infty \mathrm{d}t \, f(t) e^{-\epsilon |t|} \approx \epsilon\int_{1 / \epsilon}^0 \mathrm{d}t \, f(-\infty) + \epsilon\int_0^{1 / \epsilon} \mathrm{d}t \, f(+\infty) = f(-\infty) + f(+\infty)$

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