Exploiting symmetry in the axisymmetric Jeans equations
Background: If $f(\mathbf{x},\mathbf{v},t)$ is the distribution function of the stars in phase space and $n = \int f\,\mathrm{d}^3\mathbf{v}$ is the star density, then in the cylindrical coordinates $(R,\varphi,z)$, the $z$-component of the second of the Jeans equations applied to an axisymmetric system is
$$\renewcommand{\expv}[1]{\langle{#1}\rangle}\partial_t(n\expv{v_z}) + \partial_R(n\expv{v_Rv_z}) + \partial_z(n\expv{v_z^2}) + \frac{n\expv{v_Rv_z}}{R} + n\partial_z\Phi = 0\text{.}$$
In a steady state, the first term vanishes, so if the positive and negative terms of the $z$-component of velocity terms balance, then so do the second and fourth, in which case:
$$\partial_z(n\expv{v_z^2}) = -n\partial_z\Phi\text{,}$$
which we can then integrate.

If the the star and mass distributions are axisymmetric and symmetric about the equatorial ($z = 0$) plane, then the potential has even symmetry in $z$, $\Phi(R,z) = \Phi(R,-z)$, and so does $n$. Thus their derivatives with respect to $z$ are odd functions, and $\Phi_{,z}(R,z) = -\Phi_{,z}(R,-z)$ implies
$$\frac{1}{n}\int_{-z}^z n\Phi_{,z'}\,\mathrm{d}z' = 0\text{,}$$
since the integrand has odd symmetry. Hence,
$$\frac{1}{n}\int_z^\infty n\Phi_{,z'}\,\mathrm{d}z' = \frac{1}{n}\int_{|z|}^\infty n\Phi_{,z'}\,\mathrm{d}z'\text{.}$$