Derivation involving delta function

Question

In some of my notes on physics I am currently reading, I stumbled across something I don't understand... How do I derive$\frac{1}{\hbar^2}|\int_0^tdt'e^{i(E_n-E_m)t'/\hbar}|^2\space\space$$\rightarrow$ $\frac{1}{\hbar^2} [\frac{\sin(\omega_{nm}t/2)}{\omega_{nm}/2}]^2$ $\rightarrow$ $t\frac{2\pi}{\hbar} \delta(E_n-E_m)$ for large times. Where $\omega_{nm} = (E_n-E_m)/\hbar$ ?

Answer 1

$$ \left |\int_0^a e^{ikx} dx \right |= \left | \frac{e^{ika} - 1}{ik}\right | = \left |e^{i \frac{ka}{2}} \frac{e^{i\frac{ka}{2}} - e^{-i\frac{ka}{2}}}{ik}\right | = \left |\frac{2e^{i \frac{ka}{2}}}{k} \frac{e^{ik\frac{a}{2}} - e^{-i\frac{ka}{2}}}{2i} \right | = $$

$$\left | \frac{2 e^{i\frac{ka}{2}} }{k} \cdot \sin\left (\frac{ka}{2}\right )\right | = \left | e^{i\frac{ka}{2}} \cdot \frac{\sin \left ( \frac{ka}{2} \right )}{\frac{k}{2}}\right | = \left | \frac{\sin \left ( \frac{ka}{2} \right )}{\frac{k}{2}}\right |$$

Now substitute the right parameters and bingo.