How to evaluate this infinite sum I want to find $\int_0^1(1-x)^{\frac{1}{3}}x^{1/3}dx$.
From binomial theorem, $(1-x)^{\frac{1}{3}}= \sum_{0}^\infty (-x)^n\binom{\frac{2}{3}}{n}$. 
Then $\int_0^1(1-x)^{\frac{1}{3}}x^{1/3}dx=  \sum_{0}^\infty \frac{(-1)^n\binom{\frac{2}{3}}{n}}{n+\frac{4}{3}}$. 
i don't know how to evaluate $\sum_{0}^\infty \frac{(-1)^n\binom{\frac{2}{3}}{n}}{n+\frac{4}{3}}$ .
Thank you.

Without going through the Binomial theorem for fractional values, to can use the fact that what you want to evaluate is, up to the normalizing factor, the integral of the pdf $\beta_{\frac{4}{3}, \frac{4}{3}}$ of a $\operatorname{Beta}\!\left(\frac{4}{3}, \frac{4}{3}\right)$ distribution.. Thus, the integral you have is equal to
$$
\int_0^1 \operatorname{B}\!\left(\frac{4}{3}, \frac{4}{3}\right)\beta_{\frac{4}{3}, \frac{4}{3}}(x)dx = \operatorname{B}\!\left(\frac{4}{3}, \frac{4}{3}\right)
$$
where $\operatorname{B}\!\left(\frac{4}{3}\right) = \frac{\Gamma(4/3)\Gamma(4/3)}{\Gamma(8/3)}$ (which is in turn roughly 0.53).
