= Solution
Normalize the incident plane wave to unit amplitude. The scattering asymptotics are $\psi(\mathbf r)\sim e^{ikz}+f(\theta)e^{ikr}/r$; the outgoing coefficient $f$ is the <scattering amplitude>. The general axisymmetric free solution for $r>0$ is
$$
\psi(r,\theta)=\sum_{\ell\geq0}\{a_\ell j_\ell(kr)+b_\ell n_\ell(kr)\}P_\ell(\cos\theta),
$$
using <Spherical Bessel functions>, <spherical Neumann functions> and <Legendre polynomials>.
For a real short-range central potential, the exterior radial solution in each <partial wave> is proportional to $\cos\delta_\ell\,j_\ell-\sin\delta_\ell\,n_\ell$, whose large-$r$ form is $\sin(kr-\ell\pi/2+\delta_\ell)/(kr)$. This defines the <scattering phase shift>. Multiplying this combination by $(2\ell+1)i^\ell e^{i\delta_\ell}$ matches the incoming spherical component of the incident plane wave: the factor $e^{i\delta_\ell}$ cancels the shifted incoming phase. The outgoing coefficient instead acquires $e^{2i\delta_\ell}$. Subtracting the plane wave's outgoing component therefore gives
$$
\boxed{f(\theta)=\frac1{2ik}\sum_{\ell\geq0}(2\ell+1)(e^{2i\delta_\ell}-1)P_\ell(\cos\theta)=\frac1k\sum_{\ell\geq0}(2\ell+1)e^{i\delta_\ell}\sin\delta_\ell\,P_\ell(\cos\theta).}
$$
For the barrier's $s$-wave, the regular reduced radial function inside is $u(r)=A\sinh(\kappa r)$, with $\kappa=\sqrt{\gamma^2-k^2}$. Outside take $u(r)=B\sin(kr+\delta_0)$. Continuity of $u$ and its derivative at $a$ equates logarithmic derivatives:
$$
\kappa\coth(\kappa a)=k\cot(ka+\delta_0),\qquad\boxed{\frac{\tanh(\kappa a)}{\kappa a}=\frac{\tan(ka+\delta_0)}{ka}.}
$$
Use the branch with $\delta_0\to0$ as $k\to0$. The matching equation gives $\delta_0=k(\tanh(\gamma a)/\gamma-a)+o(k)$, so the $s$-wave <scattering amplitude> tends to
$$
\boxed{f\longrightarrow\frac{\tanh(\gamma a)-\gamma a}{\gamma}.}
$$
Its negative is the <repulsive spherical barrier scattering length>, positive for this repulsive barrier.
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