= Solution
Take $V=|\mathbf V|$ to be the test body's instantaneous <speed> relative to the nonstreaming background, rather than an ensemble <root mean square> <speed>. Use a local isotropic <galactic distribution function> $f(v)$ normalized as a number distribution, and assume independent encounters and a large, approximately constant <Coulomb logarithm in stellar dynamics>.
For a background particle with relative incoming <velocity> $\mathbf w=\mathbf V-\mathbf v$, a <gravitational scattering angle> $\chi$ obeys
$$
\tan(\chi/2)=\frac{G(M+m)}{bw^2}=\frac{b_{90}}b.
$$
In the <centre-of-momentum frame>, the test body's share of the relative <velocity> is $m/(M+m)$. The azimuthal average cancels transverse changes, leaving a change parallel to the incoming relative <velocity>,
$$
\Delta\mathbf V_\parallel=-\frac{m}{M+m}(1-\cos\chi)\mathbf w=-\frac{2m}{M+m}\frac{\mathbf w}{1+b^2/b_{90}^2}.
$$
The number of encounters per unit time in $d^3v$ and impact interval $db$ is $f(v)d^3v\,w\,2\pi b\,db$. Integrating the longitudinal change from $b=0$ to $b_m$ gives
$$
\dot{\mathbf V}=-2\pi G^2m(M+m)\int f(v)\frac{\mathbf w}{w^3}\log\!\left(1+\frac{b_m^2w^4}{G^2(M+m)^2}\right)d^3v.
$$
At leading logarithmic order replace the logarithm by $2\ln\Lambda$, evaluated at a representative <speed>. For isotropic $f$, the <velocity-shell identity for dynamical friction> is
$$
\int f(v)\frac{\mathbf V-\mathbf v}{|\mathbf V-\mathbf v|^3}\,d^3v=\frac{4\pi\mathbf V}{V^3}\int_0^Vf(v)v^2\,dv.
$$
One proof takes the divergence with respect to $\mathbf V$: the inverse-square <vector field> has divergence $4\pi\delta^{(3)}(\mathbf V-\mathbf v)$, whose velocity integral is $4\pi f(V)$, and <rotational symmetry> plus the <divergence theorem> determines the radial field. Exterior <velocity> shells give zero field. Consequently the systematic <Chandrasekhar dynamical friction> is antiparallel to motion, with
$$
\dot V=-\frac{16\pi^2G^2m(M+m)\ln\Lambda}{V^2}\int_0^Vf(v)v^2\,dv.
$$
In the massive-test-body approximation $M\gg m$, this becomes
$$
\boxed{\dot V=-\frac{16\pi^2G^2mM\ln\Lambda}{V^2}\int_0^Vf(v)v^2\,dv,\qquad \Lambda=\frac{b_mV^2}{GM}.}
$$
The <finite-test-mass dynamical friction coefficient> shows the needed qualification: $M>m$ alone does not make $M+m=M$. The quoted expression also neglects diffusion and the <velocity> dependence of the logarithmic cutoff. For the <lowered Maxwellian velocity distribution>, set $f=0$ for $v>v_e$; continuing the printed subtraction beyond escape would give a negative distribution. If $V>v_e$, the integral stops at $v_e$.
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