Derivation of GARCH Student-$t$ log-likelihood I got the general PDF of the student t distribution, that is:
$\frac{\Gamma[\frac{(\nu+1)}{2}]}{\Gamma(\frac{\nu}{2})}\,\frac{1}{\sqrt{\pi\,\nu}}\,\bigg[1 + \frac{x^2}{\nu}\bigg]^{-(\nu+1)/2}$
Bollerslev (1987) proposed a t distribution for GARCH estimation that looks like this:
$\frac{\Gamma[\frac{(\nu+1)}{2}]}{\Gamma(\frac{\nu}{2})}\,\frac{1}{\sqrt{(\nu-2)\sigma^2}}\,\bigg[1 + \frac{\epsilon_t^2}{(\nu-2)\sigma^2}\bigg]^{-(\nu+1)/2}$
I suppose that omitting $\pi$ is okay, but what I do not understand is how  and why it is possible to replace the $\nu$ with $(\nu - 2)\sigma^2$? What am I missing here?
EDIT: In order not to answer my own question, a quick edit. The solution is to consider the variance of the t-distribution, that is,
$\sigma^2 = \frac{\nu}{\nu - 2}$ and then substitute (if I got it right).
 A: For GARCH modelling with a t-distribution, we want  $y_t$ to be t-distributed with mean $\mu$ and variance $\sigma_t^2$. One way to obtain this is to consider
$$
y_t = \mu + \sigma_t \frac{1}{\sqrt{\frac{v}{v-2}}}T
$$
where T is t-distributed with $v$ degree of freedoms. Thus, 
$$
T = \frac{y_t - \mu}{\sqrt{\frac{v-2}{v}}\sigma_t}
$$
Then, replace x with the above expression in the first formula for the density of a t-distribution (put $\mu = 0$). 
A: Consider a general GARCH model:
\begin{align}
r_t&=\mu_t+\epsilon_t \\
\epsilon_t&=\sigma_t u_t
\end{align}
Usually it is assumed that $E(u_t)=0$ and $V(u_t)=1$. However, if $u_t$ follows a "normal" t-distribution with the pdf
\begin{align*}
f(u_t;\nu)=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{u_t^2}{\nu}\right)^\left(-\frac{\nu+1}{2}\right)
\end{align*}
then $E(u_t)=0$ and $V(u_t)=\frac{\nu}{\nu-2}\neq 1$ for a finite $\nu$. So the idea is to transform this density such that the variance becomes one.
To simplify notation, assume $X$ follows a "normal" t-distribution and we consider the random variabe $Y=\sigma\cdot X$ ($\sigma$ is a constant and not the variance of $X$). Then it is obvious that:
\begin{align}
E(Y)&=E(\sigma X)=\sigma E(X)=0 \\
V(Y)&=V(\sigma X)=\sigma^2V(Y)=\sigma^2\frac{\nu}{\nu-2}
\end{align}
If we choose $\sigma=\sqrt{\frac{\nu-2}{\nu}}$, we can calculate that:
\begin{align}
V(Y)=\left(\sqrt{\frac{\nu-2}{\nu}}\right)^2\frac{\nu}{\nu-2}=\frac{(\nu-2)\nu}{\nu(\nu-2)}=1
\end{align}
So the random variable $Y=\sqrt{\frac{\nu-2}{\nu}}X$ has an expected value of zero and unit variance, as desired. Now we want to derive the pdf of Y. Since $\sqrt{\frac{\nu-2}{\nu}}$ is strictly increasing, we get:
\begin{align}
f_Y(y;\nu)&=\frac{1}{\sigma}f_X\left(\frac{y}{\sigma}\right) \\
&=\frac{1}{\sigma}\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{\left(\frac{y}{\sigma}\right)^2}{\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\
&=\frac{1}{\sigma}\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\sigma^2\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\
&=\frac{1}{\sqrt{\frac{\nu-2}{\nu}}}\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\left(\sqrt{\frac{\nu-2}{\nu}}\right)^2\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\
&=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\frac{(\nu-2)\nu \pi}{\nu}} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\frac{\nu-2}{\nu}\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\
&=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{(\nu-2)\pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\nu-2}\right)^\left(-\frac{\nu+1}{2}\right) \\
\end{align}
This is the pdf of what is called a "standardized t-distribution". Because $\epsilon_t=\sigma_tu_t$, you get for the conditional distribution:
\begin{align}
f(\epsilon_t\vert {\cal F}_{t-1})=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{(\nu-2)\pi} \Gamma\left(\frac{\nu}{2}\right)}\frac{1}{\sigma_t}\left(1+\frac{y^2}{(\nu-2)\sigma_t^2}\right)^\left(-\frac{\nu+1}{2}\right)
\end{align}
However, it seems like $\pi$ is missing in your notation above and without $\pi$ this is not a valid density. Later when calculating the log-likelihood, this factor can be neglected.
