A limit and a delta function in an integral.

Question

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!

Answer 1

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 $t$.