Show that for X ~ Gamma($\alpha$, $\beta$), for positive constant $\nu$,
$E[X^\nu] = \dfrac{\beta^\nu*\Gamma(\nu + \alpha)}{\Gamma(\alpha)}$.
I have the following solution:
However, I don't understand how we get that $\dfrac{1}{\Gamma(\alpha)*\beta^\alpha}*\int_{0}^{\infty}x^{(\nu+\alpha)-1}e^{-x/\beta}dx = \dfrac{\Gamma(\nu+\alpha)*\beta^{\nu+\alpha}}{\Gamma(\alpha)*\beta^{\alpha}}$
Would appreciate any help on how this step was completed. Basically, I don't understand how: $\int_{0}^{\infty}x^{(\nu+\alpha)-1}e^{-x/\beta}dx = \Gamma(\nu+\alpha)*\beta^{\nu+\alpha}$.
I see that the left side is close to the definition of the Gamma function, but can't see how exactly to turn it into the right side.
One way to understand the calculation is to recall that for a gamma distribution with shape $\alpha$ and scale $\beta$, $$f_X(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}, \quad x > 0.$$ The denominator, being independent of $x$, suggests that $1/(\beta^\alpha \Gamma(\alpha))$ is the required multiplicative factor for the density such that $$\int_{x=0}^\infty f_X(x) \, dx = 1.$$ In other words, $$\beta^\alpha \Gamma(\alpha) = \int_{x=0}^\infty x^{\alpha - 1} e^{-x/\beta} \, dx.$$ This holds true for any $\alpha, \beta > 0$. Now, with this in mind, $$\operatorname{E}[X^\nu] = \int_{x=0}^\infty x^\nu f_X(x) \, dx = \int_{x=0}^\infty \frac{x^{\nu + \alpha - 1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \, dx = \frac{\beta^{\nu + \alpha} \Gamma(\nu + \alpha)}{\beta^\alpha \Gamma(\alpha)}\int_{x=0}^\infty \frac{x^{\nu + \alpha - 1} e^{-x/\beta}}{\beta^{\nu + \alpha}\Gamma(\nu + \alpha)} \, dx.$$ This is what the provided solution does, but the motivation for doing so should now be clear, because now the integrand represents a gamma density with shape parameter $\nu + \alpha$, and rate $\beta$. Therefore, its integral over its support is also $1$, provided $\nu + \alpha > 0$. It follows that $$\operatorname{E}[X^\nu] = \frac{\beta^\nu \Gamma(\nu + \alpha)}{\Gamma(\alpha)}, \quad \nu > -\alpha,$$ as claimed.