# Absorption probability derivation in Feynman QED

In Feynmann QED (page 11), the tansition probability

$$a_{lk} = \frac{4\sin^2(\Delta T/(2\hbar))}{\Delta^2}|u_{lk}|^2\,, \quad \Delta = E_l - E_k - \hbar\omega.$$

In the same page, there is an expression for absorption probability ($$AP$$):

$$AP = \int_0^\infty \frac{4\sin^2(\Delta T/(2\hbar))}{\Delta^2}|u_{lk}|^2 P(\omega, \theta, \phi)~d\omega~d\Omega.$$

Then it is written that when $$T$$ is large, the factor $$\frac{\sin^2(\Delta T/(2\hbar))}{\Delta^2}$$ has an appreciable value only for $$\hbar\omega$$ near $$E_l - E_k$$ and therefore $$P(\omega, \theta, \phi)$$ and $$u_{lk}$$ can be taken outside the integration. Following this, I write

$$AP = |u_{lk}|^2 P(\omega_{lk}, \theta, \phi)~d\Omega~\int_0^\infty \frac{4\sin^2(\Delta T/(2\hbar))}{\Delta^2}~d\omega$$

From here, how can I get equation (4-1):

$$\text{Trans. prob.} = \frac{2\pi}{\hbar}|u_{lk}|^2 P(\omega, \theta, \phi)~d\Omega$$

What I understood is the value of integral is $$2\pi/\hbar$$. I tried like this:

$$\because\, \Delta = E_l - E_k - \hbar\omega\,, d\omega = -d\Delta/\hbar$$

Therefore the integral is

$$I = \frac{-1}{\hbar}\int_0^\infty \frac{4\sin^2(\Delta T/(2\hbar))}{\Delta^2}~d\Delta$$

Then I substituted $$\Delta T/(2\hbar) = y$$ to get the value of integration as $$-1/\hbar \times \pi T/\hbar = -\pi T/\hbar^2$$ using

$$\int_0^\infty\frac{\sin^2y}{y^2}~dy = \pi/2$$