Assume Y has the chi-square distribution with v degrees of freedom.
I.e.
$f_Y(y) = \frac{y^{v/2 - 1}e^{-y/2}}{2^{v/2}\Gamma(v/2)}$ for $0 < y < \infty$
The fact that this is a density function implies that
$\int_{0}^{\infty} y^{v/2 - 1}e^{-y/2} = 2^{v/2}\Gamma(v/2)$. I understand everything up till here.
However, I am told that this in turn implies that for any positive number k, the mean of $Y^k$ is $\frac{2^{k}\Gamma(k+v/2)}{\Gamma(v/2)}$. How does this result arise?
$$\operatorname{E}[Y^k] = \int_{y=0}^\infty y^k f_Y(y) \, dy = \int_{y=0}^\infty \frac{y^k y^{v/2-1} e^{-y/2}}{2^{v/2} \Gamma(v/2)} \, dy = \frac{2^{v/2+k} \Gamma(v/2 + k)}{2^{v/2} \Gamma(v/2)} \int_{y=0}^\infty \frac{y^{v/2+k-1} e^{-y/2}}{2^{v/2+k} \Gamma(v/2+k)} \, dy.$$ But this last integrand is a chi-square density with $v+2k$ degrees of freedom, thus it integrates to $1$, and the expectation reduces to $$\operatorname{E}[Y^k] = 2^k \frac{\Gamma(v/2+k)}{\Gamma(v/2)}.$$
