= Solution
For azimuthally symmetric <Rutherford scattering>, integrate over the <solid angle>, $d\Omega=2\pi\sin\theta\,d\theta$. Substituting $u=\sin(\theta/2)$ gives $\sin\theta\,d\theta=4u\,du$, so
$$
\begin{aligned}
\sigma&=\frac{2\pi K}{p^2}\int_{\theta_{\min}}^{\theta_{\max}}
\frac{\sin\theta}{\sin^4(\theta/2)}\,d\theta\\
&=\frac{4\pi K}{p^2}\left[\csc^2\frac{\theta_{\min}}2-\csc^2\frac{\theta_{\max}}2\right].
\end{aligned}
$$
The small lower cutoff dominates when $\theta_{\max}\gg\theta_{\min}$ and $\theta_{\min}\ll1$. Using $\sin(\theta_{\min}/2)\simeq\theta_{\min}/2$, the leading total <cross section> is
$$
\boxed{\sigma\simeq\frac{16\pi K}{p^2\theta_{\min}^2}.}
$$
A nonzero cutoff is essential: the unscreened forward <cross section> otherwise diverges.
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