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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

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

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