= Solution
The tree-level Cooper-pair scattering amplitude is simply $V$. Define the density of states at the Fermi surface, in the normalization of the question, by
$$
\nu_F=\int_{\rm FS}\frac{d^2k}{(2\pi)^3|v_F(k)|}.
$$
In the one-loop Cooper diagram the two internal fermions have opposite momenta. After integrating over internal frequency, the remaining normal-energy integral contains
$$
\int_{|\omega|}^{\Lambda}\frac{d\xi}{\xi}
=\log\frac{\Lambda}{|\omega|}.
$$
Consequently
$$
\Gamma(\omega)=V-\nu_FV^2\log\frac{\Lambda}{|\omega|}+O(V^3\log^2),
$$
and summing the leading geometric series gives the running <BCS theory> coupling
$$
\boxed{
V_R(\omega)=\frac{V}
{1+\nu_FV\log(\Lambda/|\omega|)}.}
$$
For repulsive $V>0$ it flows logarithmically toward zero. For attractive $V<0$, its denominator vanishes at the <Cooper instability> scale
$$
\boxed{|\omega_*|=\Lambda
\exp\left[-\frac1{\nu_F|V|}\right].}
$$
At this scale the normal Fermi liquid becomes unstable: opposite-momentum fermions form Cooper pairs, a superconducting or superfluid condensate develops, and a gap opens.
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