= Solution
For a regular, spherical <star>, <hydrostatic equilibrium> gives $dP/dr=-Gm(r)\rho(r)/r^2$. Multiply by $4\pi r^3$ and integrate to the surface. <Integration by parts> gives
$$
\int_0^R4\pi r^3\frac{dP}{dr}dr=4\pi R^3P(R)-3\int_0^R4\pi r^2P\,dr.
$$
The central boundary term vanishes for finite central <pressure>. Since $dm=4\pi r^2\rho\,dr$, the right-hand side of the integrated hydrostatic equation is
$$
-\int_0^R4\pi Gm\rho r\,dr=-\int_0^M\frac{Gm}{r(m)}dm=\Omega.
$$
The last expression is the <gravitational energy>: adding a thin <mass> shell $dm$ to the interior <mass> $m$ contributes $-Gm\,dm/r$, counting each interacting pair once. Hence
$$
3\int_0^M\frac P\rho dm+\Omega=4\pi R^3P(R).
$$
With the specified zero surface <pressure> this proves the <stellar virial theorem> in the required form,
$$
\boxed{3\int_0^M\frac P\rho dm+\Omega=0.}
$$
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