The value of ${\displaystyle V_{p}}$ may also be put in the form

${\displaystyle V_{p}=a^{n}\Delta _{p}^{\frac {1}{2}}\int \ldots \int d{\frac {p_{1}}{a}}\ldots d{\frac {p_{n}}{a}}.}$

(284)

Now we may determine ${\displaystyle V_{p}}$ for ${\displaystyle \epsilon _{p}=1}$ from (279) where the limits are expressed by (281), and ${\displaystyle V_{p}}$ for ${\displaystyle \epsilon _{p}=a^{2}}$ from (284) taking the limits from (283). The two integrals thus determined are evidently identical, and we have

${\displaystyle (V_{p})_{\epsilon _{p}=a^{2}}=a^{n}(V_{p})_{\epsilon _{p}=1}}$

(285)

i. e., ${\displaystyle V_{p}}$ varies as ${\displaystyle \epsilon _{p}^{\frac {n}{2}}}$. We may therefore set

${\displaystyle V_{p}=C\epsilon _{p}^{\frac {n}{2}},\quad e^{\phi _{p}}={\frac {n}{2}}C\epsilon _{p}^{{\frac {n}{2}}-1}}$

(286)

where ${\displaystyle C}$ is a constant, at least for fixed values of the internal coördinates.

To determine this constant, let us consider the case of a canonical distribution, for which we have

${\displaystyle \int _{0}^{\infty }e^{{\frac {\psi _{p}-\epsilon _{p}}{\Theta }}+\phi _{p}}\,d\epsilon _{p}=1,}$

where

${\displaystyle e^{\frac {\psi _{p}}{\Theta }}=(2\pi \Theta )^{-{\frac {n}{2}}}.}$

Substituting this value, and that of ${\displaystyle e^{\phi _{p}}}$ from (286), we get

${\displaystyle {\frac {n}{2}}C\int _{0}^{\infty }e^{-{\frac {\epsilon _{p}}{\Theta }}}\epsilon _{p}^{{\frac {n}{2}}-1}\,d\epsilon _{p}=(2\pi \Theta )^{\frac {n}{2}},}$

${\displaystyle {\frac {n}{2}}C\int _{0}^{\infty }e^{-{\frac {\epsilon _{p}}{\Theta }}}{\bigg (}{\frac {\epsilon _{p}}{\Theta }}{\bigg )}^{{\frac {n}{2}}-1}\,d{\bigg (}{\frac {\epsilon _{p}}{\Theta }}{\bigg )}=(2\pi )^{\frac {n}{2}},}$

${\displaystyle {\frac {n}{2}}C\Gamma {\bigg (}{\frac {n}{2}}{\bigg )}=(2\pi )^{\frac {n}{2}},}$

(287)

${\displaystyle C={\frac {(2\pi )^{\frac {n}{2}}}{\Gamma ({\tfrac {1}{2}}n+1)}}.}$

Having thus determined the value of the constant ${\displaystyle C}$, we may