\[\begin{align*} E(X) &= \int_0^{\infty} x \cdot \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} dx\\ &= \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x\cdot x^{\alpha-1} e^{-\beta x} dx\\ &= \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^{\alpha} e^{-\beta x} dx\\ \therefore \text{let: } t&= \beta t \\ \therefore x &= \frac{t}{\beta} \\ \therefore dx &= \frac{dt}{\beta} \\ \therefore E(X) &= \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} \left(\frac{t}{\beta}\right)^{\alpha} e^{-t} \frac{dt}{\beta}\\ &= \frac{\beta^{\alpha}}{\Gamma(\alpha)\cdot \beta^\alpha \cdot \beta} \int_0^{\infty} t^{\alpha} e^{-t} dt\\ &= \frac{1}{\Gamma(\alpha) \beta} \cdot \Gamma(\alpha+1)\\ &= \frac{1}{\beta}\cdot\frac{\alpha\Gamma(\alpha)}{\Gamma(\alpha)} \\ &= \frac{\alpha}{\beta} \end{align*}\]
\[\begin{align*} E(X^2) &= \int_0^\infty x^2 \frac{\beta^\alpha x^{\alpha -1}e^{-\beta x}}{\Gamma(\alpha)} dx\\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha - 1} e^{-\beta x} dx \\ \textbf{let: } \beta x &= t \\ \therefore x &= \frac{t}{\beta} \\ \text{and: } dx &= \frac{dt}{\beta}\\ \therefore &= \frac{\beta^\alpha}{\Gamma(\alpha} \int_0^\infty \left(\frac{t}{\beta} \right)^{\alpha + 1} e^{-t} \frac{dt}{\beta}\\ &= \frac{\beta^\alpha}{\Gamma(\alpha) \beta^{\alpha + 2}} \int_0^\infty t^{\alpha + 1} e^{-t} dt\\ &= \frac{1}{\Gamma(\alpha) \beta^2} \int_0^\infty t^{\alpha + 1} e^{-t} dt\\ \Rightarrow \Gamma(\alpha + 2) &= \int_0^\infty t^{\alpha+1} e^{-t}dt \\ &= \frac{\Gamma(\alpha + 2)}{\beta^2 \Gamma(\alpha)} \\ &= \frac{\alpha \Gamma(\alpha + 1)}{\beta^2 \Gamma(\alpha)}\\ &= \frac{\alpha(\alpha+1) \Gamma(\alpha)}{\beta^2 \Gamma(\alpha)} \\ &= \frac{\alpha(\alpha+1)}{\beta^2}\\ Var(X) &= \frac{\alpha(\alpha+1)}{\beta^2} - \left(\frac{\alpha}{\beta}\right)^2 \\ &= \frac{\alpha^2 + \alpha}{\beta^2} - \frac{\alpha^2}{\beta^2} \\ &= \frac{\alpha^2}{\beta^2} + \frac{\alpha}{\beta^2} - \frac{\alpha^2}{\beta^2} \\ &= \frac{\alpha}{\beta^2} \end{align*}\]
\[\begin{align*} M_X(t) &= \int_0^\infty e^{tx} \frac{\beta^\alpha x^{\alpha-1}\exp\left(-\beta x\right)}{\Gamma(\alpha)} dx\\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha-1} e^{(tx-\beta x)} dx\\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty \frac{(\beta - t)^\alpha}{(\beta - t)^\alpha} x^{\alpha-1} e^{x(t-\beta)} dx \\ &= \frac{\beta^\alpha}{(\beta - t)^\alpha} \int_0^\infty \frac{(\beta - t)^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{x(t-\beta)} dx\\ & \text{and: } \int_0^\infty \frac{(\beta - t)^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{x(t-\beta)}dx = 1 \\ &= \left(\frac{\beta}{\beta - t}\right)^{\alpha} \quad \text{for } t < \beta \end{align*}\]