Suppose that $X$ follows a chi-square distribution $\chi_n^2$ and that $Y=\sqrt{2X}$. Find the pdf of $Y$ and show that $\mathbb{E}(Y) = \frac{\Gamma((n+1)/2)}{\Gamma(n/2)}$.
I have calculated the pdf using the change of variable formula as
$$f_Y(y) = \frac{y^{n-1} e^{-y^2/4}}{2^{n-1} \Gamma(n/2)}$$
Now $$ \begin{split} \mathbb{E}(Y) &= \int^{\infty}_{0} y \cdot \frac{y^{n-1} e^{-y^2/4}}{2^{n-1} \Gamma(n/2)}dy \\ &= \int^{\infty}_{0} \frac{y^{n} e^{-y^2/4}}{2^{n-1} \Gamma(n/2)}dy \\ &= \int^{\infty}_{0}\frac{y^{n-1}e^{-y^2/4}}{2^{n-2}\Gamma(n/2)} \frac{y}{2}dy \\ &= \int^{\infty}_{0}\frac{2^{n-1}(y^2/4)^{(n-1)/2}e^{-y^2/4}}{2^{n-2}\Gamma(n/2)} \frac{y}{2}dy \\ &= \int^{\infty}_{0}\frac{2(y^2/4)^{(n+1)/2-1}e^{-y^2/4}}{\Gamma(n/2)} \frac{y}{2}dy. \end{split} $$
Finally using the definition that $\Gamma(\alpha) = \int^{\infty}_{0} x^{\alpha-1}e^{-t}dt$ and making the substitution $t=y^2/4 \Leftrightarrow dt = y/2\,dy$ we get
$$ \mathbb{E}(Y) = \frac{2}{\Gamma(n/2)} \int^\infty_0 t^{(n+1)/2-1}e^{-t}\,dt = \frac{2\Gamma((n+1)/2)}{\Gamma(n/2)} $$ which is off by a factor of $2$ of what they get. Where is the mistake?