Integrating out $\sigma^2$ My question is with regards to integrating on $\sigma^2$ when working with normal distributions. How does one handle the $^2$ aspect of $\sigma^2$?
I am unsure whether I should treat $\sigma^2$ as an entity in itself (in other words, the "squared" aspect is just part of the symbol) and integrate with respect to $d\sigma^2$ or if I need to integrate with respect to $d\sigma$ (and treat the exponent in the usual way, and use substitution if necessary). 
I believe that the answer to my question is that it depends on what information I have about $\sigma$.  For example, if I have a $PDF$ for $\sigma^2$ and I want to find the $CDF$ I would integrate with respect to  $d\sigma^2$, not $d\sigma$
In a slightly more complicated scenario, what if I have a distribution for $\sigma$ and a distribution for $X|\sigma^2$ and I would like to find the marginal distribution for $X$. ie: 
$\int f_X(x|\sigma^2)f_\sigma(\sigma) ..d?$
Edit: 
Come to think of it, if I have something like $\sigma^2$ ~ $U(a, b)$ would I write that as
$f_\sigma = ... $ or would I write $f_{\sigma^2} = ...  $   (again, my intuition tells me the latter). 

Assume that $\sigma^2$ has density $g$ and that, conditionally on $\sigma=s$, $X$ has density $f_s$. In other words, for every measurable $A$ and $B$,
$$
\mathbb P(\sigma^2\in A)=\int_Ag(r)\mathrm dr,\qquad\mathbb P(X\in B\mid \sigma=s)=\int_Bf_s(x)\mathrm dx.
$$
Then, for every $s\geqslant0$,
$$
\mathbb P(\sigma\geqslant s)=\mathbb P(\sigma^2\geqslant s^2)=\int_{s^2}^{+\infty}g(r)\mathrm dr,
$$
hence the density of $\sigma$ is $h$, where
$$
h(s)=-\frac{\mathrm d}{\mathrm ds}\int_{s^2}^{+\infty}g(r)\mathrm dr=2sg(s^2),
$$
and $X$ has (unconditional) density $f$, where
$$
f(x)=\int_0^{+\infty} f_s(x)h(s)\mathrm ds=\int_0^{+\infty}  f_s(x)2sg(s^2)\mathrm ds=\int_0^{+\infty}  f_{\sqrt{r}}(x)g(r)\mathrm dr.
$$
