= Solution
Set $V=\Delta\dot\alpha$, positive downwards. Then $h_t=-V\cos\theta$. The leading axisymmetric <lubrication> flux is $q_\theta=-h^3p_\theta/(12\mu a)$. Wall Couette flux from the translating sphere is smaller by $\Delta/a$ than the squeeze flux and does not affect this leading <pressure>. The spherical <Reynolds lubrication equation> is
$$
h_t+\frac1{a\sin\theta}\partial_\theta(\sin\theta\,q_\theta)=0.
$$
Regularity at the pole gives $q_\theta=(aV/2)\sin\theta$, so
$$
p_\theta=-\frac{6\mu a^2V\sin\theta}{\Delta^3(1-\alpha\cos\theta)^3},\qquad
\boxed{p=\frac{3\mu a^2V}{\alpha\Delta^3(1-\alpha\cos\theta)^2}+p_*}.
$$
The apparent singularity at $\alpha=0$ is a constant-pressure gauge term: subtract $3\mu a^2V/(\alpha\Delta^3)$ before taking the limit. The remaining <pressure> tends to $6\mu a^2V\cos\theta/\Delta^3$.
To leading lubrication order the <force> is the <pressure> integral. Its downward component on the inner sphere is
$$
F_z=-2\pi a^2\int_0^\pi p\cos\theta\sin\theta\,d\theta
=-\frac{6\pi\mu a^4V}{\alpha^3\Delta^3}\int_{-\alpha}^{\alpha}\frac{t}{(1-t)^2}\,dt.
$$
The original PDF's first integration hint is incorrect: the rational term needs $1-b^2$, not $(1-b)^2$. From the antiderivative $1/(1-t)+\log(1-t)$,
$$
\int_{-b}^b\frac{t}{(1-t)^2}\,dt=\frac{2b}{1-b^2}+\log\frac{1-b}{1+b}.
$$
Thus the <squeeze resistance of eccentric nested spheres> gives
$$
\boxed{F_z=-\frac{6\pi\mu a^4V}{\alpha^3\Delta^3}\left[\frac{2\alpha}{1-\alpha^2}+\log\frac{1-\alpha}{1+\alpha}\right].}
$$
It opposes the <velocity> for either sign of $\alpha$. The continuous concentric limit is $F_z=-8\pi\mu a^4V/\Delta^3$, since the bracket is $4\alpha^3/3+O(\alpha^5)$. As $\alpha\to1$, its leading magnitude is $6\pi\mu a^4|V|/[\Delta^3(1-\alpha)]$.
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