In Born approximation the differential cross section is given by \[\begin{equation} \frac{d\sigma}{d \Omega}=\left| \frac{m}{2 \pi}\int d^3r e^{-i {\bf q\cdot\, r }} V_I(r)\right|^2,\tag{1} \end{equation}\] where \(V_I(r)=e V(r)\) is the interaction energy of the incoming electron with the hydrogen atom. It will be better to do the calculation in two parts since \(V(r)\) is contributed by two charge constituents, the proton and the electron. The contribution from the proton is straightforward to compute since we know \(V_p(r)\) right away: \(V_p(r)=\frac{e}{r}\). The corresponding integral to be calculated as \[\begin{equation} I_1\equiv\frac{m}{2 \pi}\int d^3r e^{-i {\bf q\cdot\, r }}\frac{e^2}{r}.\tag{2} \end{equation}\] Unfortunately this integral is divergent, but we can regularize it by introducing an exponentially decaying function \(e^{-\epsilon r}\), with \(\epsilon>0\). What we do is to evaluate the integral with \(V_p(r)=\frac{e^2 e^{-\epsilon r}}{r}\) and let \(\epsilon\rightarrow 0\) in the final answer (unlike physicists, mathematicians would strongly object: “You cannot change the order of integration and the limit \(\epsilon\rightarrow0\)”. We can think of it as assigning a finite number to a divergent integral by the renormalization procedure. Physically \(e^{-\epsilon r}/r\) terms represents the potential due a massive particle of mass \(\epsilon\).) In this case we have \[\begin{eqnarray} I_1&\equiv &\frac{m e^2}{2 \pi}\int d^3r e^{-i {\bf q\cdot r }}\frac{ e^{-\epsilon r}}{r}=2 m e^2 \int dr \frac{\sin (q r)}{q}e^{-\epsilon r}=\frac{2 m e^2}{q} \Im \left\{ \int_0^\infty dr e^{-r(\epsilon -i q)}\right\}\nonumber\\ &=&\frac{2 m e^2}{q} \Im\left\{ \frac{1}{\epsilon -i q}=\frac{2 m e^2}{\epsilon^2+ q^2}\right\}=\frac{2 m e^2}{q^2},\tag{3} \end{eqnarray}\] where we finally set \(\epsilon=0\).
The second contribution comes from the electron cloud. The potential due to electron in the ground state is given as \[\begin{eqnarray} V_e(r)&=&\ e(\frac{ e^{- r/a_0}}{r}-\frac{1}{ r}+\frac{1}{a_0} e^{- r/a_0}).\tag{4} \end{eqnarray}\] The first idea would be to plug this into the integral, and have fun… It can be done, but let’s try something else, \[\begin{eqnarray} I_2&\equiv &\frac{m e}{2 \pi}\int d^3r e^{-i {\bf q\cdot \,r }} V_e(r) = \frac{2 m e}{q}\int_0^\infty r V_e(r)\sin (q r)dr\tag{5} \end{eqnarray}\] Now the idea is the integration by parts twice with the definitions \({\mathcal U}=r V(r)\) an \(d{\mathcal V}= \sin (q r)\) for the first one and \({\mathcal U}=\frac{d}{d r}(r V(r))\) an \(d{\mathcal V}= \cos (q r)\) for the second one. Note that we can drop \({\mathcal U} {\mathcal V}|_0^\infty\) since \({\mathcal U}\) vanishes at the boundaries. After two integration by parts we get \[\begin{eqnarray} I_2&=& -\frac{2 m }{q^3}\int_0^\infty \frac{d^2}{d r^2}(r V_e(r))\sin (q r)dr.\tag{6} \end{eqnarray}\] What we have achieved is \(\frac{d^2}{d r^2}(r V_e(r))\) term, which is r \(\nabla^2 V_e\). By Poisson’s equation it can be replaced by \(r (-\,4 \pi\rho_e(r))\) where \[\begin{equation} \rho_e(r)=-e |\psi(r)|^2= -\frac{e}{\pi a_0^3}e^{-2 r/{a_0}}\tag{7} \end{equation}\] So the integral we need to deal with is \[\begin{eqnarray} I_2&=& -\frac{8 m e^2}{a_0^3 q^3}\int_0^\infty r e^{-2 r/{a_0}}\sin (q r)dr=-\frac{8 m e^2}{a_0^3 q^3}\Im\left\{ \int_0^\infty r e^{-(2 /{a_0}-i q) r}dr\right\}\nonumber\\ &=&-\frac{8 m e^2}{a_0^3 q^3}\Im\left\{ \left(\frac{-d}{d \alpha}\right)_{\alpha=(2 /{a_0}-i q)}\int_0^\infty e^{-\alpha r}dr\right\}\nonumber\\ &=&-\frac{8 m e^2}{a_0^3 q^3}\Im\left\{ \frac{q a_0}{(2-i q a_0)^2}\right\}=-\frac{32 m e^2}{ [4+(q a_0)^2]^2 q^2}.\tag{8} \end{eqnarray}\] Combining both terms we have \[\begin{eqnarray} \frac{d\sigma}{d\Omega}&=&\frac{4 m^2 e^4}{q^4}\left(1-\frac{16}{ [4+(q a_0)^2]^2 }\right)^2.\tag{9} \end{eqnarray}\] Before we discuss the limiting cases, let’s solve the problem one more time using a shortcut. The object we are dealing with is, \[\begin{equation} I=\int d^3r e^{-i {\bf q\cdot\, r }} V(r),\tag{10} \end{equation}\] which is nothing but the Fourier transform of the potential energy. Instead of jumping onto the problem head on, let’s follow a detour. Consider the Poisson’s equation, \[\begin{equation} \nabla^2 V(r)= -4 \pi \rho(r)=-4\pi e(\delta (\overrightarrow{r})-\rho_e(r)).\tag{11} \end{equation}\] Instead of Fourier transforming the potential itself, we can Fourier transform Eq. (11) which gives \[\begin{equation} -q^2 \tilde{V}(q)=-4\pi e(1-{\cal F}\{\rho_e(r)\})\,\rightarrow\,\tilde{V}(q)=\frac{4\pi e(1-{\cal F}\{\rho_e(r)\})}{q^2}.\tag{12} \end{equation}\] \({\cal F}\{\rho_e(r)\}\) term is to be calculated by usual means, and in reproduces the second term in Eq (9).
For large \(q a_0\) the second term in Eq. (9), which accounts for the electron cloud, can be neglected. So the incoming electrons are deflected only by the proton. For small \(q a_0\) the cross section becomes zero which is a manifestation of the fact that the hydrogen atom is neutral. This means that only highly deflected electrons can probe into the electron cloud, and “see” the proton.