Let $U \sim N(0,1)$ and $V \sim \chi_{p}^{2}$. $U$ and $V$ independent. Set $\displaystyle T = \frac{U}{\sqrt{\frac{V}{p}}}$.
What I want to show is $$f_T(t) = \frac{\Gamma((p+1)/2)}{\sqrt{\pi p}\Gamma(p/2)} \left(1+\frac{t^2}{2}\right)^{-(p+1)/2},$$ for $-\infty < t < \infty$.
I first set W = V. Find the inverse transformations $h_1^{-1}$ and $h_2^{-1}$ and the Jacobian of the transformation.
$h_1^{-1}(t, W) = V = t\sqrt{W/p} $ and $h_2^{-1}(t, W) = V = W$
Jacobian is $\sqrt{W/p}$
Then I find the joint density of T and W $$f_{T, W}(t,W) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(t\sqrt{W/p}^2} \frac{W^{p/2 - 1} e^{-w/2}}{2^{p/2}\Gamma(p/2)}\sqrt{\frac{W}{p}}$$ since U and V are independent
However, when I integrate out W to find the marginal density of T...I set $$f_T(t) = \frac{1}{\sqrt{2\pi}} \frac{1}{2^{p/2}\Gamma(p/2)} \int_0^\infty e^{\frac{t^2w^2}{4p}} w^{p/2-1} \sqrt{\frac{w}{p}} \, dw$$...I do not how to evaluate the thing inside the integral