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 $0 < \beta < 1$?
 A: The antiderivative can be expressed in terms of the Whittaker M function or incomplete Gamma function.  But to solve for $\alpha$ will certainly require numerical methods.
A: in general, even for simple $\beta$ (like $\beta=2$) there is no analytic anti-derivative. You could expand this into power series and integrate within the radius of convergence, but that would also likely place you in the realm of numerical methods.
A: Actually, I think (as it was earlier mentioned) that it is not possible to express it in closed analytic form (only numeric solution), but for some cases, for example for $\beta=2$ one can do the following. 
 Set $t=(\frac{2}{\sigma})^{\frac{1}{\beta}}x$, then: 
$$P = \frac{1}{\sigma}\displaystyle\int_{0}^{\alpha} \exp (- \frac{ 2 x^{\beta}}{\sigma} ) \;dx=\frac{\sigma^{\frac{1}{\beta}-1}}{2^{\frac{1}{\beta}}}\displaystyle\int_{0}^{(\frac{2}{\sigma})^{\frac{1}{\beta}}\alpha} \exp (- t^{\beta}) \;dx=\frac{1}{\sqrt{2\sigma}}\displaystyle\int_{0}^{\sqrt{\frac{2}{\sigma}}\alpha} \exp (- t^{2})\;dx$$
$$P=\frac{1}{\sqrt{2\sigma}}\frac{\sqrt{\pi}}{2}\text{erf}\left(\sqrt{\frac{2}{\sigma}}\alpha\right)$$
Then $$\alpha=\sqrt{\frac{\sigma}{2}}\text{erf}^{-1}\left(\frac{2\sqrt{2\sigma}P}{\sqrt{\pi}}\right).$$
Where $\text{erf}^{-1}$ - is the inverse error function.
UPDATE
I think one can find analytic solution. Not pure analytic but rather “quasi-analytic” if it is possible to say so.
Changing the variable $t= Bx^{\beta}, \quad \mathrm{d}x=\mathrm{d}t\frac{B^{-\frac{1}{\beta}}}{\beta }t^{\frac{1}{\beta}-1}$ yields:
$$P = \frac{1}{\sigma} \displaystyle\int_{0}^{\alpha} \exp (- B x^{\beta}) \;dx=\frac{ B^{-\frac{1}{\beta}}}{\sigma \beta}\displaystyle\int_{0}^{B\alpha^{\beta}} e^{-t} t^{\frac{1}{\beta}-1}\mathrm{d}t $$
Using the definition of the lower incomplete gamma function:
$$ \gamma(s,x) = \int_0^x t^{s-1}\,e^{-t}\,{\rm d}t .\,\!$$
$$P=\frac{ B^{-\frac{1}{\beta}}}{\sigma \beta}\gamma\left(\frac{1}{\beta}, B\alpha^{\beta}\right) $$
Assuming that many CAS systems have algorithms for numeric inversion of gamma function (and its variations like incomplete gamma functions) (for example you can use MatLAB) one can obtain the solution in the following manner:
$$\alpha=\left(  \frac{1}{B}\gamma^{-1}\left(\frac{1}{\beta},\beta\sigma PB^{\frac{1}{\beta}} \right)  \right)^\frac{1}{\beta}$$
which is valid for arbitrary positive values of $P,\sigma,B,\beta$.
A: 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.
