Solve for $\alpha$: $P = \frac{1}{\sigma}\int_{0}^{\alpha} \exp (\frac{ -2 x^{\beta}}{\sigma} ) dx$ I need to solve:
$$P = \frac{1}{\sigma}\int_{0}^{\alpha} \exp ( \frac{ -2 x^{\beta}}{\sigma} ) \;dx $$
This simplifies to:
$$P = \frac{1}{\sigma} \int_{0}^{\alpha} \exp (- B x^{\beta}) \;dx $$
But if we let:
$$t^{2} = Bx^{\beta}$$
And try to make it an erf, then:
$$2t dt = \beta B x^{\beta-1} dx$$
This could continue ad-infinitum. Any ideas? Or is a numerical solution the only thing we can do?
Update: So for $\beta = 2$, an analytical solution has been found. What about the case for $0 < \beta < 1$?

As Robert Israel answered, the result is given in terms of the incomplete gamma function. This can can also simplify to $$P = \frac{1}{\sigma}\displaystyle\int_{0}^{\alpha} \exp \Big(\frac{ -2 x^{\beta}}{\sigma}\Big ) dx=-\frac{\alpha }{\beta  \sigma }\,E_{\frac{\beta -1}{\beta }}\left(\frac{2 \alpha ^{\beta }}{\sigma }\right)$$ (provided $\Re(\beta )>0$) where appears the exponential integral function.
This is a very complex function of $\alpha$ and, as already answered, only numerical methods could solve the problem.
