I'm trying to integrate this function on Wolfram Alpha, but it says "no result found in terms of standard mathematical functions".
$$\frac{\alpha t^{\alpha-1}e^{-(t/\beta)^\alpha}}{\beta^\alpha}$$
However, if I just substitute $u$ for $-(t/\beta)^\alpha$, I'd get the integral as $-e^{-(t/\beta)^\alpha} + C$.
I'm not sure what I'm doing wrong here. Could the community please help me figure out my mistake?
Thanks.
Let $u = (t/\beta)^\alpha.$ Then $du = \alpha(t/\beta)^{\alpha-1}(dt/\beta).$ Therefore $$ \int \frac{ \alpha t^{\alpha -1} e^{-(t/\beta)^\alpha}} {\beta^\alpha} \, dt = \int \alpha \left( \frac t \beta \right)^{\alpha-1} e^{-(t/\beta)^\alpha} \left( \frac {dt} \beta\right) = \int e^{-u^\alpha} \, du. $$ In any course on theory of probability you hear that this has no closed form when $\alpha = 2.$ I would guess it has a closed form only for special values of $\alpha,$ perhaps only for $\alpha=0$ and $\alpha=1.$