How to simplify out of the gamma function $\Gamma(z)$? $$\mathrm{d}t(x_i; \nu) = \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2}) \sqrt{\pi\nu}} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} } \enspace, i=1,2$$
How to derive a closed-form solution of the above by simplifying out of the gamma function?
Is the gamma function that is being used above (for $z>0$) the same as
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\mathrm{d}x ?$$
 A: The symbol $\Gamma$ in the factor  $\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2}) \sqrt{\pi\nu}} $ does denote the gamma function.
The expression for the pdf of the univariate Student's t-distributution $dt(x_i; \nu)$ is already in closed-form in the sense that the gamma function with an arbitrary real argument cannot be expressed in terms of more elementary functions other than in integral, infinite series, or infinite product representations (many of which are shown in the link). I don't believe that would qualify as "simplifying out of the gamma function".
If the argument is a positive integer, then we have the immense simplification $\Gamma(n) = (n-1)!$. However, for a degree-of-freedom parameter $\nu \not \in \mathbb{Z}$, both arguments $\nu/2$ and $(\nu+1)/2$ are not integers. Even if $\nu$ is an integer one or the other of the arguments is not an integer.
An alternative representation that "appears" less complicated uses the beta function which is related to the gamma function through
$$\int_0^1 t^{x-1}(1-t)^{y-1} \, dt := B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Hence,
$$\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\sqrt{\pi}}= \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\Gamma (\frac{1}{2})}= \frac{1}{B \left(\frac{1}{2}, \frac{\nu}{2} \right)},$$
and $\displaystyle dt(x_i;\nu) = \frac{1}{\sqrt{\nu}B \left(\frac{1}{2}, \frac{\nu}{2} \right)} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} }.$

One may ask why the gamma function appears in this pdf to begin with. A random variable $X$ that is Student's t-distributed with $\nu$ degrees of freedom is obtained as $X = Z/S$ where $Z \sim \mathcal{N}(0,1)$ is a standard normal random variable and $S = \sqrt{Y/\nu}$ where $Y \sim \chi^2(\nu)$ has a chi-square distribution.
The cdf for $Y$ is
$$F_Y(y) = C \int_0^y u^{(\nu-2)/2}e^{-u/2} \, du,$$
where  the normalizing factor $C$ is chosen so that $F_Y(y) \to 1$ as $y \to \infty$, that is
$$\frac{1}{C}= \int_0^\infty u^{(\nu-2)/2}e^{-u/2} \, du = \int_0^\infty 2^{\nu/2 -1}t^{\nu/2 -1 }e^{-t} \, 2dt = 2^{\nu/2} \Gamma(\nu/2).$$
After many more steps of transformation the pdf $dt(x;\nu)$ is obtained, retaining the factor $\Gamma(\nu/2)$ and introducing $\Gamma((\nu+1)/2)$.
