Finding the expectation of the Gamma density function 
I am confused on how to solve this problem. I understand that there is some relationship along the lines of: 
$$ \Gamma(\alpha) = \int e^{(-t)}t^{\alpha -1} $$ 
$$ \Gamma(\alpha) = \int e^{-x/\beta}(x/\beta)^{\alpha -1} $$ 
$$ \Gamma(\alpha) = 1/(\beta)^{(\alpha-1)}\int e^{-x/\beta}(x)^{\alpha-1} $$ 
$$ (\beta)^{(\alpha-1)}\Gamma(\alpha) = \int e^{-x/\beta}(x)^{\alpha-1} $$ 
So the function then simplifies to: 
$$ 1/\beta $$
Which would be the expectation. However the answer given is $$\alpha\beta $$ 
So I am not entirely sure what I am doing wrong. 
Thanks! 
 A: Let $k > 0$.  Then $$\begin{align*}
\operatorname{E}[X^k] &= \int_{x=0}^\infty x^k f(x;\alpha,\beta) \, dx \\
&= \int_{x=0}^\infty x^k \frac{x^{\alpha-1} e^{-x/\beta}}{\Gamma(\alpha)\beta^\alpha} \, dx \\
&= \int_{x=0}^\infty \frac{\Gamma(\alpha+k) \beta^k}{\Gamma(\alpha)} \cdot \frac{x^{\alpha+k-1} e^{-x/\beta}}{\Gamma(\alpha+k) \beta^{\alpha+k}}  \, dx \\
&= \frac{\Gamma(\alpha+k)\beta^k}{\Gamma(\alpha)} \int_{x=0}^\infty f(x;\alpha+k,\beta) \, dx.
\end{align*}$$  Since $f$ is a probability density for any positive choice of parameters, this last integral is equal to $1$; therefore, $$\operatorname{E}[X^k] = \frac{\Gamma(\alpha+k) \beta^k}{\Gamma(\alpha)}.$$  When $k$ is a positive integer, $\Gamma(\alpha+k) = (\alpha+k-1)(\alpha+k-2)\ldots \alpha \Gamma(\alpha)$ by the recursion relation, thus we would have $$\operatorname{E}[X^k] = \prod_{m=0}^{k-1} (\alpha+m).$$  When $k = 1$, this simply yields $\Gamma(\alpha+1) = \alpha \Gamma(\alpha)$ hence $$\operatorname{E}[X] = \alpha \beta$$ as claimed.  Our more general approach furnishes the higher raw moments, e.g., $$\operatorname{E}[X^2] = (\alpha+1)\alpha \beta^2$$ which also gets us the variance $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \alpha \beta^2.$$
