How to compute this form factor? [closed]

Question

I would really appreciate if someone could help me about computing the form factor $$F(q^2)=\int_0^{\infty}\rho(r)e^{-iq\cdot r}d^3r \quad, $$ when $$\rho (r)=\begin{cases}\rho_0& r<R\\0& r>R\end{cases}\quad . $$ The answer is $$F(q^2)=\frac{3(\sin (qr)-qr\cos(qr))}{(qr)^3},$$ where $r=|\overset{\to}{r}|$, $q=|\overset{\to}{q}|$, $\overset{\to}{q}$, and $\overset{\to}{r}$ are three dimensional vectors, but I don't know how to obtain it. Thanks in advance.

What I've tried: Using spherical coordinates, we have $d^3 r=dxdydz=r^2dr\sin \theta d\theta d\phi$, where $0\leq r\leq R$, $0\leq \theta \leq \pi$, and $0\leq \phi \leq 2\pi$. Then $$\int_{0}^{\infty}e^{-i\overset{\to}{q}\cdot \overset{\to}{r}}d^3 r=\int\int\int e^{-irq\cos\theta}r^2dr\sin \theta d\theta d\phi$$ $$=2\pi \int\int e^{-irq\cos\theta}r^2dr\sin \theta d\theta .$$ Now how can I compute the last integral?

Answer 1

You can use the fact $$ \int e^{-iqr \cos\theta} \sin \theta d\theta = -\int e^{-iqr c} dc $$ where $c$ is the integration variable that replaced $\cos\theta$. The integral is now elementary. The form factors for spheres (this case), rods and slabs are reported in, e.g., this paper (see equation 42). For more details, see also Scattering from homogeneous objects and this answer.