Proving that $\int_{-\infty}^\infty f(x) \, dx = 1$ , where $f(x)$ is the density of Student t distribution How do i prove the following:
$$
 \int_{-\infty}^\infty  f(x) \, dx = \frac{\Gamma \left( \frac{\nu + 1}{2}
  \right)}{\sqrt{\nu \pi}\, \Gamma \left( \frac{\nu}{2} \right)} \int_{-\infty}^\infty \left( 1 + \frac{x^2}{\nu} \right)^{-(\nu + 1)/2} \, dx =1
$$ 
where $f(x)$ represents the density of Student t - distribution.
 A: We have:
\begin{align}&\int_{-\infty}^\infty \left(1 + \frac{x^2}{\nu}\right)^{-(\nu+1)/2}\, dx\\
&= 2\int_0^\infty \left(1 + \frac{x^2}{\nu}\right)^{-(\nu+1)/2}\, dx\\
&= 2\int_0^\infty (1 + y)^{-(\nu+1)/2} (1/2)\nu^{1/2}y^{-1/2}\, dy \quad (\text{substitute $y = x^2/\nu$})\\
&= \nu^{1/2}\int_0^\infty y^{1/2 - 1}(1 + y)^{-(\nu + 1)/2}\, dy\\
&= \nu^{1/2} B\left(\frac{1}{2}, \frac{\nu + 1}{2} - \frac{1}{2}\right) = \nu^{1/2}B\left(\frac{1}{2}, \frac{\nu}{2}\right) \\
&= \nu^{1/2}\Gamma\left(\frac{1}{2}\right)\frac{\Gamma\left(\frac{\nu}{2}\right)}{\Gamma\left(\frac{\nu}{2} + \frac{1}{2}\right)}\\
&= \sqrt{\nu\pi}\frac{\Gamma\left(\frac{\nu}{2}\right)}{\Gamma\left(\frac{\nu + 1}{2}\right)}.
\end{align}
Hence 
$$ \frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu\pi}\Gamma\left(\frac{\nu}{2}\right)}\int_{-\infty}^\infty \left(1 + \frac{x^2}{\nu}\right)^{-(\nu + 1)/2}\, dx = 1.$$
Note: I used the facts $\Gamma(1/2) = \sqrt{\pi}$, $$B(x,y) = \int_0^\infty t^{x - 1}(1 + t)^{-x-y}\, dt \quad (x > 0, y > 0)$$
and 
$$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} \quad (x > 0, y > 0).$$
