Rutherford's $\alpha$-Particle Scattering Experiment In Ernest Rutherford's  $\alpha$-particle scattering experiment, it is well-known that the solid angle density of  $\alpha$-particle flux
$$\frac{d\dot{N}}{d\Omega}$$
on the inner surface of a sphere of which the centre is the point of impingement of a beam of  $\alpha$-particles on a (extremely) thin gold foil is proportional to 
$$\csc^4\frac{\phi}{2}$$
where $\phi$ is the angle away from the pole of the apparatus - the point on the sphere upon which an undeflected beam would impinge. This proportionality relation incurs the problem of having no normalisation: integration of it over even the smallest non-zero region of the sphere comprising the pole will yield an infinite flux. 
But there are various mechanisms operating to blur the distribution of flux over the sphere and 'wash-out' the singularity at the pole, such as the non-zero width of the beam and the non-zero spread of directions in the beam. Clearly, at the pole flux-density is not going to exceed the flux density of the beam, and 
$$\frac{d\dot{N}}{d\Omega}\leq \frac{a^2\dot{N_0}}{\pi b^2}$$
where $\dot{N_0}$ is the flux in the beam, $a$ is the radius of the sphere, & $b$ the radius of the beam. 
I do not find in Rutherford's paper that there is any attempt to hardwire this blurring mathematically into the proportionality relation - it probably wasn't necessary to do so in order sufficiently to evince that the nucleus is an extreme concentration of charge and mass; and besides, the detection medium at the point on the sphere antipodal to the $\alpha$-particle source was almost certainly utterly saturated: but I am curious anyway as to how it would be hardwired in. I fairly sure the greatest contribution to the blurring would be the finite width of the beam, followed by the spread of directions in the beam. I think the contribution from the finity of cross-section about a nucleus, in that more than a certain distance away another nucleus would be encountered, would be a small one. 
So the question is primarily "how would the finite cross-sectional area of the beam be hardwired into the expression for solid-angle-density of flux (the $\csc^4\frac{\phi}{2}$ expression) so that it can be normalised?". It's almost certainly going to be an integration over a disc centred on point at polar-angle $\phi$ (I know $\phi$ normally denotes azimuth these days, but Rutherford used it for polar-angle); but I can't quite catch the particular details of how to do it.
Postscript
It's also quite remarkable, I think, the sheer serendipity of  how in those days they just barely had the resources to do this experiment: if you calculate what proportion of the flux must be collimated out to get a beam reasonably narrow and sufficiently close to thoroughly parallel, and factor in that $\alpha$-particles cannot escape from deep within a bulk of substance, it transpires that radium has just sufficient activity to make the experiment feasible within a reasonable time span. These days the experiment could be reproduced in a trice with a piece of polonium-210, or one of many other nuclides.
 A: Let $r$ be radius within the $\alpha$-particle beam away from the axis of the beam; and let $b$ be the closest approach to the nucleus of an $\alpha$-particle travelling straight at it.
$$r=\frac{b}{2}\cot\frac{\phi}{2}$$
Introduce a 'unit' sigmoid function (ie has unit gradient at origin, is odd, and tends to ±1 as its argument tends to $\pm\infty$) $\digamma$ that 'confines' $r$ to $r\leq a$. This could be $\arctan x$ or $\tanh x$ or $\dfrac{x}{\sqrt{1+x^2}}$, or anything atall that fits the sigmoidity critærion; and set 
$$r=a\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right).$$
This basically 'squashes' the whole arrangement into a circle of radius $a$, with $a$ acting as a parameter.
Also let $\dot{N}$ denote flux & $\dot{n}$ denote flux density & $\dot{n}_0$ be the $\alpha$-particle flux density in the beam, & $\dot{N}_0$ be the total $\alpha$-particle flux in the beam. We have then 
$$\frac{d\dot{N}}{d\phi}
=
\frac{\dot{n}_0\pi ab}{2}\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\digamma'\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\csc^2\frac{\phi}{2}$$
It can be seen that if $\digamma(x)$ tends to the horizontal line $y=\pm1$ even so slow as $x^{-\epsilon}$ (epsilon an arbitrarily small positive number, then the index of the resulting function as $\phi\rightarrow 0$ will be $-1+\epsilon$, by reason of $\digamma' ×\csc^2$, whence the total function integrable.
(It looks kind of odd that the expression explicitly in terms of $\digamma$ & its derivative should have a value independent of the precise nature of  $\digamma$ ... but that becomes clear by noting that the integral is simply of $r\cdot dr$; also that insofar as it itself is smaller it's derivative is greater, & vice versa, over most of its range.)
And then
$$\begin{align}\frac{d\dot{N}}{d\Omega}&=\frac{d\dot{n}}{2\pi\sin\phi d\phi}=\frac{d\dot{n}}{4\pi\sin\frac{\phi}{2}\cos\frac{\phi}{2} d\phi}
\\&=\frac{\dot{n}_0\pi ab}{8}\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\digamma'\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\sec\frac{\phi}{2}\csc^3\frac{\phi}{2}
\\&=
\frac{\dot{N}_0\pi b}{8a}\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\digamma'\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\sec\frac{\phi}{2}\csc^3\frac{\phi}{2}\end{align}$$
applying the normalisation, which is now an elementary matter - no integration needs to be done because we have effectively done the derivation by reversing the normalisation integral.
By expressing it using the total $\alpha$-particle flux density, it covers the scenario of there being many nuclei instead of just one - the many 'discs' each of radius $a$ behaving like just one 'disc with all the flux going through it ... except insofar as many take up space laterally! The matter of how the non-zero radius - indeed many atoms radius - acts to blurr the beam. I still haven't sorted that ... but what I have done here is to introduce a roughly plausibly physical tweak to the distribution - plausible in that an $\alpha$-particle with its trajectory departing laterally from a nucleus will be caught up eventually in the field of another at a distance from the first that corresponds to the parameter $a$ - the radius of the disc within which, in the above derivation, the field is slightly distorted to fall to zero at the edge. If $a$ be larger by a substantial ratio than  $b$, which is physically realistic, as in reality $b$ is much less than the semi-interatomic separation ... so the distribution will be pretty much the same as before - just now normalised. Speaking qualitatively, the reason the distribution in its untwoken form has a singularity (and a ^4 one at that!) is that there is in (growing) infinitude of lateral displacement corresponding to $\phi\rightarrow 0$.
@Batominovski
I haven't as yet tried (to completion) that method with the Yukawa potential. I found that when I did try it, and commenced the derivation for $r$ in terms of $\phi$, I got an integral I couldn't solve. Even for the case of a simple Coulomb potential, I had to wheel-out Gradstein & Ryzhik! The integral necessary for the solution using Yukawa potential might be buried in there somewhere ... but I couldn't find it in the time. The Wolfram Integrator gave up the ghost! 
But then ... I might be using an unnecessarily hard method for solving for the deflection in terms of perpendicular distance of nucleus from undeviated path of $\alpha$-particle (isn't "impact parameter" the correct name for that?) - I do that sort of thing sometimes. If anyone knows a simple method, please tell me!
But the Yukawa-potential method looks very similar to mine in many respects, with its single best-fit parameter.
And that 
$$(\frac{2mV_0}{\hbar^2\mu})^2\frac{1}{(2k^2(1-\cos\theta)+\mu^2)^2}$$
is exactly the kind of expression I was looking for. But I haven't as yet figured-out the precise mathematical machinery for actually obtaining it! Also, I notice it removes the pole in the density (solid angle density $d\dot{N}/d\Omega$) distribution itself. My solution for that is still singular (or rather can still be - this does depend on the choice of $\digamma$), and depends on the factor $2\pi\sin\phi$ (still  clinging  to Rutherford's notation!) that arises in integration over a sphere for integrability.
Actually, looking it up, they don't even do it, the classical way atall. Doing it the classical way, conserving momentum & energy about the nucleus, the integral is gotten
$$\int\frac{dr}{r\sqrt{r^2 -br\exp(-kr)-a^2}}$$
, with $a$ the impact parameter, is obtained. I cannot find this in Gradsteyn & Ryzhyk (how many "i"s & "y"s are there in the correct spelling of that!?), & the Wolfram Online Integrator reports a "Computation time exceeded!" error. It's fine - though a tad awkward - with $k=0 \therefore$ the exponential factor $=1$ in it.
Also the $b$ in this is not the same $b$ as in the non-Yukawa-ised form of this this problem & simply brought over from  it: rather $b$ will now be given by a lambertw function ... but that's not so very bad atall. Infact, it'll just be 
$$\frac{w(kb_0)}{k} ,$$
with $b_0$ being the $b$ in the non-Yukawa-ised form. The closest approach of the α-particle being well within the shielding corresponds to $kb_0$ being a small fraction of 1, & therefore $b$ being not much less than $b_0$ ... $b≈b_0(1-kb_0(1-\frac{3}{2}kb_0(1-\frac{16}{9}kb_0)))$, infact.
Does anyone by any chance know how to do the special "f" for "function" in LateX? It's character 133 in ASCII.
