= Solution
The asymptotic <scattering wavefunction> is
$$
\boxed{\psi(\mathbf r)\sim e^{ikz}+f(\theta)\frac{e^{ikr}}r.}
$$
In each <partial wave>, the <Partial-wave S-matrix> is the ratio of its outgoing coefficient to its incoming coefficient. With the conventions in the question,
$$
S_l=1+2if_l.
$$
For <elastic scattering> by a real central potential, <unitary operator>[unitarity] gives $|S_l|=1$, so $S_l=e^{2i\delta_l}$ for a real <scattering phase shift> $\delta_l$. Hence
$$
\boxed{f_l=\frac{S_l-1}{2i}=e^{i\delta_l}\sin\delta_l.}
$$
Using the <Orthogonality of Legendre polynomials> in $\sigma_T=\int|f(\theta)|^2d\Omega$ yields the <partial-wave total scattering cross-section>
$$
\boxed{\sigma_T=\frac{4\pi}{k^2}\sum_{l=0}^\infty(2l+1)\sin^2\delta_l.}
$$
For the <S wave>, write the reduced radial wavefunction as
$$
u_{<}(r)=A\sinh(\kappa r),
\qquad
u_{>}(r)=B\sin(kr+\delta_0),
\qquad
\kappa=\sqrt{\gamma^2-k^2},
$$
where regularity at $r=0$ selected the hyperbolic sine. Continuity of $u$ and $u'$ at $r=a$, equivalently <logarithmic-derivative matching>, gives
$$
k\cot(ka+\delta_0)=\kappa\coth(\kappa a),
$$
or
$$
\boxed{\frac{\tan(ka+\delta_0)}{ka}
=\frac{\tanh(\sqrt{\gamma^2-k^2}\,a)}{\sqrt{\gamma^2-k^2}\,a}.}
$$
For $ka\ll1$, let the <scattering length from a partial-wave S-matrix>[scattering length] be $a_s=-\lim_{k\to0}\tan\delta_0/k$. Expanding the matching relation gives
$$
a-a_s=\frac{\tanh(\gamma a)}\gamma,
\qquad
a_s=a-\frac{\tanh(\gamma a)}\gamma.
$$
Only the S wave contributes at leading order, so
$$
\boxed{\sigma_T^{(0)}\sim4\pi
\left(a-\frac{\tanh(\gamma a)}\gamma\right)^2.}
$$
In the <hard-sphere limit> $\gamma a\to\infty$, penetration is suppressed, $a_s\to a$, and
$$
\boxed{\sigma_T^{(0)}\to4\pi a^2.}
$$
This is four times the geometric area $\pi a^2$, a wave effect associated with diffraction from an impenetrable sphere.
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