I am trying to use this equation
$Prob_d(\chi^2 > \chi_0^2) = \frac{2} {2^{d/2} \Gamma(d/2)} \int_{\chi_0}^{\inf} x^{d-1}e^{-x^2/2} dx$
to compute probabilities of an empirical distribution $\chi^2_0$ ($d$ degrees of freedom). The equation is from Taylor 1997: An Introduction to Error Analysis.
(How) Is this expression related to the $\chi^2$ cumulative density distribution?
$p=\frac{1}{2^{d/2} \Gamma(d/2)}\int_{0}^{x}t^{d/2-1}e^{-t/2}dt$
Can I use available numerical implementations of the $\chi^2$ CDF to compute the former?
The equation can be transformed using a substitution.
Write the first equation as $$P = \frac{2} {2^{d/2} \Gamma(d/2)} \int_{\chi_0}^{\infty} x^{d-1}e^{-x^2/2} dx$$
Define $t=x^2$, so $dt=2xdx =2t^{1/2}dx$
And so $dx=\frac{1}{2}t^{-1/2}dt$
Substituting we obtain $$P = \frac{2} {2^{d/2} \Gamma(d/2)} \int_{\chi_0^2}^{\infty} {(t^{1/2})}^{d-1}e^{-t/2}. \frac{1}{2}t^{-1/2}dt$$
Combining exponents of $t$ and taking the $1/2$ out we obtain
$$P=\frac{1}{2^{d/2} \Gamma(d/2)}\int_{\chi_0^2}^{\infty}t^{d/2-1}e^{-t/2}dt$$
as expected.
I am not sure however if I have treated the integration limits correctly.