I met a problem in the book by John R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions. In the question 13.4 (page:258), I need to integrate:
$$ \int^{\infty}_0r^2dr|\psi(\vec{x},t)|^2 $$
, thereof $\psi$ (the scattered wave) is: $$ |\psi(\vec{x},t)|^2=2\pi m\Gamma^2|Y^0_{l}(\hat{x})|^2|\phi_l(E_R)|^2\frac{e^{-\Gamma (t-r/v_R)}}{r^2p_R} \theta(t-\frac{r}{v_R}) $$
,where $m$ is the mass of the incident particle, $\Gamma$ is the imaginary term of the resonance energy, $Y^0_l(\hat{x})$ is the spherical harmonic function and $\hat{x}$ means the polar angle $\theta$ and azimuthal angle $\phi.$ $E_R=\frac{p^2_R}{2m}$ is the resonance energy. $v_R=\frac{p_R}{m}$ is the velocity of the scattered particle at resonance. $t$ is the observation time. $\theta(x)$ is a unit Heaviside function.
($m,\ \Gamma,\ p_R,\ v_R$ are constants)
Then, after the integral I need to show that the result is:
$$ \frac{2\pi^2(2l+1)}{p_R^2}\Gamma |Y^0_l(\hat{p})|^2 m p_R\int d\Omega|\phi(\vec{p})|^2 $$ , where $l>0$ means the angular momentum of the resonance scattered particle. $\phi(\vec{p})$ is the incident wave function and is well peaked along the z-axis. $\Omega$ is the solid angle. $\hat{p}$ is the solid angle of the incident wave.
The expansion of $\phi(\vec{p})$ in terms of spherical harmonic functions contains only $m=0$ terms because it is well peaked along the z-axis.
$$ \phi(\vec{p})=\sum_l\phi_l(E)Y^0_l(\hat{p}) $$
I tried my best to calculate it out but I failed.
I don't know how many physics students will see this problem but it might be the place here to post my problem.
Thank you