# Median of beta distribution for alpha = beta

I've seen that the median of the Beta Distribution cannot be defined by a closed form analytic expression but in most sites I see people give that, when the parameters $$\alpha$$ and $$\beta$$ are the same, the median is $$1/2$$. Moreover, I've seen other cases such as $$\text{median} = 1 - 2^{-1/\beta}$$ whenever $$\alpha = 1$$ and $$\beta > 0$$, but I am not being able to get these results by working out this:

$$\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \Gamma(\beta)} \int_0^{m(\alpha,\beta)} x^{\alpha-1} (1-x)^{\beta-1} \, dx = \frac12.$$

Can someone help me to work out the expression to get a closed formula for these cases?

The case $$\alpha = \beta$$ is easy since the PDF is symmetric about the point $$x=1/2$$.
For $$\alpha = 1$$ and $$\beta > 0$$, note that $$\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} = \beta$$ so $$\int_0^m f(x)\,dx = \beta\int_0^m (1-x)^{\beta-1} \, dx = [-(1-x)^\beta ]_{x=0}^m = 1 - (1-m)^\beta.$$ Setting this equal to $$1/2$$ yields $$m = 1-2^{-1/\beta}$$.