Calculating pdf, expectation and variance of random variable

Question

Suppose that X follows a chi-square distribution $\chi_n^2$ and that $Y=\sqrt{2X}$. Find the pdf of $Y$ and show that $\mathbb{E}(Y) = \frac{\Gamma((n+1)/2)}{\Gamma(n/2)}$ and $\mathbb{E}(Y^2) = 2n$.

I have calculated the pdf using the change of variable formula as $f_Y(y) = y^{n-1} e^{-y^2/4}/(2^{n-1} \Gamma(1/2n))$. However how would i calculate $\mathbb{E}(Y)$ and $\mathbb{E}(Y^2)$? Is there a trick to calculating the integral

$\mathbb{E}(Y) = \int^{\infty}_{0} \frac{y e^{-y^2/4}}{2^{n-1}\Gamma(1/2n)} dy$ as it looks messy?

My guess would be $\mathbb{E}(Y^2)$ would follow from a similar calculation?

Answer 1

$\mathbb{E}(Y^2) = \mathbb{E}(2 \cdot X) = 2 \cdot \mathbb{E}(\chi^2_{n}) = 2 \cdot n$

Answer 2

Integral of type $\int\limits_{0}^{+ \infty} x^p \cdot e^{-x^2} \ dx$ is computed easily by making a substitution $\{ x^2 = t \}$ and making use of Euler Gamma function $\{ \Gamma(\alpha) = \int\limits_{0}^{+ \infty} x^{\alpha - 1} \cdot e^{-x} \ dx \}$