\[\begin{align*} E(X) &= \int_0^\infty x \frac{k}{\lambda} \left( \frac{x}{\lambda} \right)^{k-1} e^{-(x/\lambda)^k} dx \\ &= k \int_0^\infty \left( \frac{x}{\lambda} \right)^{k} e^{-(x/\lambda)^k} dx \\ \text{let: } u &= \left(\frac{x}{\lambda}\right)^k \\ \Rightarrow x &= \lambda \sqrt[k]{u} \\ dx &= \frac{\lambda}{k}u^{(\frac{1-k}{k})}du \\ E(X) &= k \int_0^\infty ue^{-u} \frac{\lambda}{k}u^{(\frac{1-k}{k})}du\\ &= \lambda\int_0^\infty u^{(\frac{1-k}{k})+1}e^{-u} du \\ &= \lambda\int_0^\infty u^{(\frac{1}{k})+1-1}e^{-u} du \\ &=\lambda \Gamma\left(1 + \frac{1}{k}\right) \end{align*}\]
\[\begin{align*} E(X^2) &= \int_0^\infty x^2 \frac{k}{\lambda} \left( \frac{x}{\lambda} \right)^{k-1} e^{-(x/\lambda)^k} dx \\ &= k \int_0^\infty x\left(\frac{x}{\lambda} \right)^{k} e^{-(x/\lambda)^k} dx \\ \text{let: } u &= \left(\frac{x}{\lambda}\right)^k \\ \Rightarrow x &= \lambda \sqrt[k]{u} \\ dx &= \frac{\lambda}{k}u^{(\frac{1-k}{k})}du \\ E(X^2) &= k \int_0^\infty \lambda u^{\frac{1}{k}} u e^{-u} \frac{\lambda}{k}u^{(\frac{1-k}{k})}du\\ &= \lambda^2 \int_0^\infty u^{(\frac{1}{k} + 1 + \frac{1-k}{k})}e^{-u}du\\ &= \lambda^2 \int_0^\infty u^{(\frac{1 + k + 1-k}{k})}e^{-u}du\\ &= \lambda^2 \int_0^\infty u^{(\frac{2}{k})}e^{-u}du\\ &= \lambda^2 \int_0^\infty u^{(\frac{2}{k})+1-1}e^{-u}du\\ &= \lambda^2 \Gamma\left(1+\frac{2}{k}\right) \\ \mathrm{Var}(X) &= \lambda^2 \Gamma\left(1+\frac{2}{k}\right) -\left(\lambda\Gamma\left(1 + \frac{1}{k}\right)\right)^2\\ &= \lambda^2 \left[\Gamma\left(1 + \frac{2}{k}\right) - \left(\Gamma\left(1 + \frac{1}{k}\right)\right)^2\right] \end{align*}\]
\[\begin{align*} M_X(t) &= \int_0^\infty e^{tx}\frac{k}{\lambda} \left( \frac{x}{\lambda} \right)^{k-1} e^{-(x/\lambda)^k} dx \\ &= \frac{k}{\lambda} \int_0^\infty e^{tx} \left(\frac{x}{\lambda} \right)^{k-1} e^{-(x/\lambda)^k} dx \\ &= \frac{k}{\lambda} \int_0^\infty \left(\frac{x}{\lambda} \right)^{k}\left(\frac{x}{\lambda} \right)^{-1} e^{tx} e^{-(x/\lambda)^k} dx \\ \text{let: } u &= \left(\frac{x}{\lambda}\right)^k \\ \Rightarrow x &= \lambda \sqrt[k]{u} \\ dx &= \frac{\lambda}{k}u^{(\frac{1-k}{k})}du \\ M_X(t) &= \frac{k}{\lambda} \int_0^\infty u \cdot u^{-\frac{1}{k}} e^{t\lambda u^{1/k}} e^{-u} \frac{\lambda}{k}u^{(\frac{1-k}{k})}du \\ &= \int_0^\infty u^{1-\frac{1}{k} + \frac{1-k}{k}} e^{t\lambda u^{1/k}} e^{-u}du \\ &= \int_0^\infty u^{\frac{k-1+1-k}{k}{k}} e^{t\lambda u^{1/k}} e^{-u}du \\ &= \int_0^\infty e^{t\lambda u^{1/k}} e^{-u}du \\ & \text{By definition of the Maclaurin Series:} \\ &= \sum_{n=0}^\infty \frac{(\lambda t)^n}{n!} \int_0^\infty u^{\frac{n}{k}}e^{-u} du \\ &= \sum_{n=0}^\infty \frac{(\lambda t)^n}{n!}\Gamma(\frac{n}{k} +1), \text{ for } k \ge 1 \end{align*}\]