Is this relation correct? How do I get there? It's about the following relation:
$$
\biggl|\int_{-T/2}^{T/2}e^{-it(E_p-E_k)/\hbar}\,dt\biggl|^2=2\pi\hbar T\delta(E_p-E_k) \textrm{,   for $T\rightarrow\infty$ }
$$
This is out of a QM course discussing the scattering of two particles in the Yukawa potential. The mentioned relation was used and I really don't see why it is equal. I know that $\int_{-\infty}^{\infty}e^{itx}\,dt=2\pi\delta(x)$ and $\delta(ax)=\frac{1}{\vert a\vert}\delta(x)$. This is somewhat close to the relation but I am clueless about the rest.
 A: We can start by computing the integrals on the left-hand side:
$$ \int \limits_{-T/2}^{T/2} \mathrm{e}^{\pm \mathrm{i} (E_p - E_k)t/\hbar} \mathrm{d} t = \frac{\sin(T (E_p-E_k) /2\hbar)}{(E_p-E_k)/2\hbar} $$
(for $E_k = E_p$ the result is $T$, which is just the limit of the right-hand side as $E_k \to E_p$). Therefore, we can rewrite your statement as
$$ \lim_{T \to \infty} \frac{4 \hbar^2 \sin^2(T (E_p-E_k) /2\hbar)}{T(E_p-E_k)^2} = 2 \pi \hbar \delta(E_p - E_k) \, . $$
This is a limit in the sense of distributions, so we need to show that both sides give the same result when applied to an arbitrary test function (a Schwartz function $f \in \mathcal{S}(\mathbb{R})$, for example). We know that
$$ \int \limits_{\mathbb{R}} 2 \pi \hbar \delta(E_p - E_k) f(E_k) \, \mathrm{d} E_k = 2 \pi \hbar f(E_p) \, . $$
For the left-hand side we let $T (E_p - E_k)/2\hbar = x$ to obtain
\begin{align}
\lim_{T \to \infty} \int \limits_{\mathbb{R}} \frac{4 \hbar^2 \sin^2(T (E_p-E_k) /2\hbar)}{T(E_p-E_k)^2} f(E_k) \, \mathrm{d} E_k &= \lim_{T \to \infty} 2 \hbar \int \limits_{\mathbb{R}} f\left(E_p - \frac{2 \hbar x}{T}\right) \frac{\sin^2(x)}{x^2} \, \mathrm{d} x \\
&=  2 \hbar \int \limits_{\mathbb{R}} \lim_{T \to \infty} f\left(E_p - \frac{2 \hbar x}{T}\right) \frac{\sin^2(x)}{x^2} \, \mathrm{d} x
\\
&= 2 \hbar f(E_p) \int \limits_{\mathbb{R}} \frac{\sin^2(x)}{x^2} \, \mathrm{d} x = 2 \pi \hbar f(E_p) \, ,
\end{align}
where we used the dominated convergence theorem to take the limit inside the integral. The value of the final integral is derived here.
