= Solution
Assume a stationary, spherical, nonrotating <collisionless stellar system>, a constant <mass-to-light ratio> $\Upsilon$, and vanishing boundary stresses at infinity. Let $\lambda$ be the <luminosity density>, so $\rho=\Upsilon\lambda$, and set $\gamma=G\Upsilon$. Distinguish spatial radius $r$ from projected radius $R$. The enclosed <luminosity> and radial force are
$$
L(r)=4\pi\int_0^r\lambda(s)s^2\,ds,\qquad \Phi'(r)=\frac{GM(r)}{r^2}=\frac{\gamma L(r)}{r^2}.
$$
Here $\sigma_t^2$ is the <velocity dispersion> of either one of the two tangential components; thus $\beta=1-\sigma_t^2/\sigma_r^2$. If a combined tangential variance is used instead, its contribution must be divided by two in this definition.
To derive the <Jeans equation>, start with the steady <Collisionless Boltzmann equation> $v_j\partial_j f-(\partial_j\Phi)\partial_{v_j}f=0$. Multiply by $v_i$ and integrate over all velocities. <Integration by parts> in velocity, with $f$ decaying fast enough to eliminate boundary terms, gives
$$
\partial_j\bigl(\lambda\overline{v_iv_j}\bigr)=-\lambda\partial_i\Phi.
$$
The same equation holds with luminosity weighting because the tracer's luminosity is conserved along its orbit. Spherical symmetry makes the local <stellar velocity moment> tensor diagonal, with entries $q=\lambda\sigma_r^2$, $q_t=\lambda\sigma_t^2$, $q_t$. Its radial <divergence> is $q'+(2q-2q_t)/r$: the second term accounts for the changing directions of the radial and tangential unit vectors. Therefore
$$
\boxed{\frac{d(\lambda\sigma_r^2)}{dr}+\frac{2\beta}{r}\lambda\sigma_r^2=-\frac{\gamma\lambda L(r)}{r^2}.}
$$
This is the <Spherical Jeans equation>; it follows from a <stellar velocity moment> rather than requiring a particular form of the <galactic distribution function>.
Integrating the <luminosity density> along a line of sight gives its <surface brightness>:
$$
\mu(R)=2\int_R^\infty\frac{\lambda(r)r\,dr}{\sqrt{r^2-R^2}},\qquad \lambda(r)=-\frac1\pi\int_r^\infty\frac{\mu'(R)\,dR}{\sqrt{R^2-r^2}}.
$$
The second identity is <Spherical Abel deprojection>. For the second <stellar velocity moment>, the angle between the radial direction and the line of sight has $\sin^2\theta=R^2/r^2$. The line-of-sight <velocity dispersion> is therefore $\sigma_r^2\cos^2\theta+\sigma_t^2\sin^2\theta=\sigma_r^2(1-\beta R^2/r^2)$, and
$$
S(R)\equiv\mu(R)\sigma_p^2(R)=2\int_R^\infty\left(1-\beta(r)\frac{R^2}{r^2}\right)\frac{q(r)r\,dr}{\sqrt{r^2-R^2}}.
$$
From the <Spherical Jeans equation>, $\beta q=-rq'/2-\gamma\lambda L/(2r)$. Substitution produces
$$
S(R)=2\int_R^\infty\frac{q(r)r\,dr}{\sqrt{r^2-R^2}}+R^2\int_R^\infty\frac{q'(r)\,dr}{\sqrt{r^2-R^2}}+\gamma R^2\int_R^\infty\frac{\lambda(r)L(r)\,dr}{r^2\sqrt{r^2-R^2}}.
$$
The primary PDF has a genuine symbol error in its projected gravity term: it prints a second $\gamma$ multiplying $L$ inside the integral, where $\lambda$ is required. The corrected <projected spherical Jeans residual> is
$$
\boxed{p(R)=S(R)-\gamma R^2\int_R^\infty\frac{\lambda(r)L(r)\,dr}{r^2\sqrt{r^2-R^2}}=2A(R)+RA'(R),\quad A(R)=\int_R^\infty\frac{q(r)r\,dr}{\sqrt{r^2-R^2}}.}
$$
To justify the derivative at the singular-looking lower limit, write $A(R)=\int_0^\infty q(\sqrt{R^2+z^2})\,dz$. Differentiation then gives $A'(R)=R\int_R^\infty q'(r)(r^2-R^2)^{-1/2}dr$. In particular, $Rp=(R^2A)'$. If $R^2A$ tends to zero both at zero and infinity, as for a finite isolated equilibrium with sufficiently decaying stresses,
$$
\int_0^\infty 6\pi R p(R)\,dR=6\pi[R^2A(R)]_0^\infty=0.
$$
If a boundary stress survives, its contribution must be retained; the zero-integral statement is not a finite-aperture identity.
Now identify each integral with a term in the <virial theorem>. Let $K$ be the total <kinetic energy>, including all three random components, and $W$ the <gravitational potential> energy. Averaging over spatial directions in a spherical system gives one third of the total <stellar velocity moment> along any chosen line of sight, even if the velocities are locally anisotropic. Thus
$$
2\pi\int_0^\infty R S(R)\,dR=\int\lambda\overline{v_{\rm los}^2}\,d^3x=\frac{2K}{3\Upsilon},\qquad K=3\pi\Upsilon\int_0^\infty R S(R)\,dR.
$$
Building the spherical mass shell by shell gives
$$
W=-\int_0^\infty\frac{GM(r)}{r}\,dM(r)=-4\pi G\Upsilon^2\int_0^\infty r\lambda(r)L(r)\,dr=-4\pi\gamma\Upsilon\int_0^\infty r\lambda L\,dr.
$$
Finally, <Fubini theorem> and $\int_0^r R^3(r^2-R^2)^{-1/2}dR=2r^3/3$ give the gravitational contribution to the residual integral explicitly:
$$
6\pi\gamma\int_0^\infty R^3dR\int_R^\infty\frac{\lambda(r)L(r)\,dr}{r^2\sqrt{r^2-R^2}}=4\pi\gamma\int_0^\infty r\lambda L\,dr=-\frac{W}{\Upsilon}.
$$
Consequently
$$
\boxed{6\pi\int_0^\infty Rp(R)\,dR=\frac{2K+W}{\Upsilon},\qquad \gamma=\frac{3\int_0^\infty R\sigma_p^2(R)\mu(R)\,dR}{2\int_0^\infty r\lambda(r)L(r)\,dr}.}
$$
The <global projected virial estimate of a mass-to-light ratio> is independent of the <velocity-anisotropy parameter> under these assumptions. It does not remove the <mass--anisotropy--density degeneracy> for arbitrary radial mass profiles. In particular, if an independent <dark matter halo> makes mass fail to follow light, a single constant $\Upsilon$ is no longer the physical enclosed mass-to-light ratio at every radius.
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