Let's say I want to calculate the integral \begin{equation} \lim_{\gamma\to0}\frac{1}{2\pi i}\int^\infty_{-\infty}dx\left(\frac{1}{x-a+i\gamma}-\frac{1}{x-a-i\gamma}\right)\frac{-1}{e^x+1}, \end{equation} ($a\in\mathbb{R}$) and somebody says to me that the relation \begin{equation} \frac{1}{\pi}\lim_{\gamma\to0}\frac{\gamma}{x^2+\gamma^2}=\delta(x) \end{equation} might be useful. Rewriting the fractions in the integrand I write the integral as \begin{equation} \lim_{\gamma\to0}\frac{1}{\pi}\int^\infty_{-\infty}dx\,\frac{\gamma}{(x-a)^2+\gamma^2}\frac{1}{e^x+1}, \end{equation} and being willing to use the given relation I might want to continue to write \begin{equation} \int^\infty_{-\infty}\frac{1}{\pi}\lim_{\gamma\to0}\frac{\gamma}{x^2+\gamma^2}\frac{1}{e^x+1}=\frac{1}{e^a+1}, \end{equation} and be happy because this is the answer we are supposed to find. However, I'm not sure as to why this works like this (if it does). I know that interchanging limits and integrals can be a delicate business, and if you can pull the limit inside the integral here, then what is wrong with writing \begin{equation} \lim_{\gamma\to0}\left(\frac{1}{x-a+i\gamma}-\frac{1}{x-a-i\gamma}\right)=\frac{1}{x-a}-\frac{1}{x-a}=0? \end{equation} I'd be happy with any help!

  • $\begingroup$ First, it's true that $$\lim_{\gamma \to 0} \left( \frac{1}{x-a + i \gamma} - \frac{1}{x-a-i\gamma} \right) = \frac{1}{x-a} - \frac{1}{x-a} = 0$$ But it's FALSE that $$\lim_{\gamma \to 0} \int_\Bbb{R} \left( \frac{1}{x-a + i \gamma} - \frac{1}{x-a-i\gamma} \right) =\int_\Bbb{R} \lim_{\gamma \to 0} \left( \frac{1}{x-a + i \gamma} - \frac{1}{x-a-i\gamma} \right)$$ As you said : tricky business. $\endgroup$ – Tryss Jan 8 '17 at 20:47
  • $\begingroup$ Thanks for confirming this. Do you have some insight as to how to approach the integral? $\endgroup$ – B. Pasternak Jan 8 '17 at 20:57
  • $\begingroup$ It's even tricker than than you know; when doing this sort of calculation, some of those things that look like limits of functions are meant to be limits of different sorts of things, and some of those $\int$-expressions aren't integrals of functions! $\endgroup$ – Hurkyl Jan 8 '17 at 21:06
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    $\begingroup$ @B.Pasternak : To do this properly, I would use complex analysis. If you integrate the function $\frac{1}{z-a} \frac{ 1}{e^z +1}$ on the rectangle $M+i\gamma, -M-i\gamma$, with the residue theorem, you should get the answer (disclaimer : I didn't do the calculations) $\endgroup$ – Tryss Jan 8 '17 at 21:29
  • $\begingroup$ @Tryss Integrating on that rectangle might not give the desired result, since by my estimate, you would end up with $1/(e^{x \pm i\gamma} + 1)$ in the integrand instead of $1/(e^x + 1)$. $\endgroup$ – Joshua Mundinger Jan 9 '17 at 0:02

There are a variety of ways to make the details precise, and the theory has a number of awkward points, but here is a sketch of how things typically go. See wikipedia for more details.

There is some algebra $\mathcal{T}$ of "test functions" which includes $-\frac{1}{e^x + 1}$.

There is some space $\mathcal{F}$ of functions with the proprerty that, for every $f \in \mathcal{F}$, we can define a linear functional

$$ I_f : \mathcal{T} \to \mathbb{C} : t \mapsto \int_{-\infty}^{\infty} f(x) t(x) \, \mathrm{d} x $$

and whenever $f \neq g$, there exists a test function $t$ such that $I_f(t) \neq I_g(t)$.

Finally there is some dual space $\mathcal{T}^*$ consisting of all linear functionals on $\mathcal{T}$ of an appropriate type.

What's happening is that if we have a family $f_\gamma$ of functions in $\mathcal{F}$, we can compute

$$ \lim_{\gamma \to 0} \int_{-\infty}^{\infty} f_\gamma(x) t(x) \, \mathrm{d}x = \lim_{\gamma \to 0} I_{f_\gamma}(t) $$

Now, for every $g \in \mathcal{F}$, we have $I_g \in \mathcal{T}^*$. The type of dual space we use is selected so that evaluation is continuous; that is, we can compute the above by first computing the limit in $\mathcal{T}^*$:

$$ \lim_{\gamma \to 0} I_{f_\gamma}(t) = \left( \lim_{\gamma \to 0} I_{f_\gamma} \right)(t)$$

The thing you were told is that if we set $$ f_\gamma(x) = \frac{\gamma}{x^2 + \gamma^2} $$ then $$ \frac{1}{\pi} \lim_{\gamma \to 0} I_{f_\gamma} = \delta $$

and so putting everything together gives

$$ \lim_{\gamma \to 0} \int_{-\infty}^{\infty} f_\gamma(x) t(x) \, \mathrm{d}x = \pi t(0) $$ for every test function $t$.


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