can anyone suggest how to solve this intergral? $$\int_{0}^\infty\alpha\Big(\frac{x}{\beta}\Big)^{\alpha}e^{-\left(\frac{x}{\beta}\right)^\alpha}dx,~~\alpha>0,~~\beta>0$$.
I'm not familiar with it, I guess it will involve integral by parts with substitution.
Let $$ t=(\frac{x}{\beta})^{\alpha}\implies x =\beta t^{1/\alpha}$$ that is $$dx = \frac1\alpha \beta t^{1/\alpha -1} dt$$ Hence, we have: $$\int_{0}^\infty\alpha(\frac{x}{\beta})^{\alpha}e^{-(\frac{x}{\beta})^\alpha}dx= \int_{0}^\infty te^{-t}\beta t^{1/\alpha -1} dt\\=\beta \int_{0}^\infty e^{-t}t^{1+\frac{1}{\alpha} -1} dt=\beta \Gamma\left(\frac{1}{\alpha}+1\right)$$
Where $$\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt,~~~x>0$$