Based on my understanding, the differential cross section $\frac{d\sigma}{d\Omega}$ gives us the differential area of incident particles that projects onto a differential piece of solid angle on the detector. When one calculates total cross section, one integrates over the entire solid angle; i.e. $\sigma = \int d\sigma = \int \frac{d\sigma}{d\Omega}d\Omega$. I do not comprehend this. The total cross section is indicative of the likelihood of an interaction between the incident beam and the target, and let us assume that the incident beam is a beam ranging across all impact parameters. If at impact parameter large enough the incident particles do not deflect and thus has a scattering angle of close to zero, wouldn't there be an unbounded range of impact parameter scattering onto near zero angles, and thus any integration that passes through zero will result in a blow up? (not just for rutherford scattering)
[Physics] Interpretation of total cross section in scattering
scattering-cross-section
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In a lot of ways what follows is only a partial answer. I think you are getting confused because you are trying to reason about infinities using the same tools you would apply to numbers.
I want to start with a set of practical considerations.
Where in the classical definition are the zero-angle deflected particles not counted in the scattering cross section?
The classical description includes particles that are scattered through a vanishingly small angle, but no real experiment need worry about those for two reasons:
The equipment include both beam-pipe and detector elements has finite size. As does the prepared beam. There is always an lower limit on the angle at which scattered and un-scattered particles can be distinguished. The result is that the integration of the cross-section should not be taken to cover $4\pi$.
The target generally does not consist of a single scattering center, but of a macroscopic quantity of matter meaning that the beam particles are in principle scattered by many centers. In most cases the measured scattering is dominated by a single hard (or at least harder) scattering event. In any case, scattering at impact parameters above about half the inter-center distance is mostly averaged out (lookup "multiple scattering" for the statistical properties of this process).
This leads to a practical notion that particles are either scattered or not scattered, and the scattering cross-section concerns itself only with the former. Now, that generates some tension with the understanding that the scattering cross-section is infinite for long-range forces. (I think this is the source of your question.)
You can dodge the tension by several means:
Noting that measured scattering cross-sections are always averaged over finite chunks of solid angle, and can't distinguish between no interaction and very little interaction. That is, the infinite value for long ranges forces is a black-board physics construct (albeit a deeply useful one).
By making a semantic distinction between "infinitesimally scattered" and "not scattered". The divergent part of the long-range cross-section goes in the former category, which the "missed the effective range" part of the short-range cross-section goes in the latter category not-withstanding that they are experimentally indistinguishable.
Scattering cross sections can have other parameters besides angle. For example you commonly have cross sections vs. angle and final energy, $d\sigma/d\Omega\, dE_\text{final}$. This might reflect the fact that in a generic elastic scattering process forward-scattered particles tend to retain most of the beam energy, while backwards-scattered particles must deposit a lot of energy and momentum in the target.
If you're doing purely elastic scattering from a crystal, for example Bragg scattering, the cross section $d\sigma/d\Omega$ will essentially vanish except for some very particular angles. It's the values of those magic angles that are useful, and the relative strengths of scattering at one angle versus another. The fact that the entire incident beam has to go somewhere can sometimes be a useful check, but more often the unscattered / forward-scattered beam gets sent to some beam dump and ignored.
Best Answer
It should be noted that the differential cross section at $\theta = 0$ is not really defined - what is the difference between a particle being scattered by angle zero and not being scattered at all? However, it's completely legitimate to ask about the behavior of $\frac{d\sigma}{d\Omega}$ as $\theta$ approaches close to zero.
For Rutherford scattering, the total cross-section blows up as we integrate $\frac{d\sigma}{d\Omega}$ closer and closer to $\theta=0$. This is because the scattering probability decreases too slowly as the impact parameter grows, so the contribution from larger and larger impact parameters cannot be neglected (the integral diverges as $b \to \infty $). Fundamentally, this is because the photon is massless, so the EM interaction strength falls as $1/r^2$ (known as a long-range interaction).
For interactions mediated by massive particles, the interaction strength falls as $\frac{e^{-mr}}{r^2}$ (known as a short-range interaction). There is a non-zero scattering probability for every impact parameter, but this probability decays exponentially, so the integral converges at $b \to \infty$.
For classical hard-sphere scattering, the situation is even simpler - the scattering probability actually hits zero for impact parameters above a threshold. Therefore the integral also converges as $b \to \infty$.