Let x ~ Beta (2,$\frac{1}{2}$). Then calculate $E\left(\frac{2}{1+x}\right)$.
So, ${E}[g(X)] = \displaystyle \int_{-\infty}^\infty g(x) f(x)\, \mathrm{d}x$ .
$\displaystyle f(x;\alpha,\beta) =\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\, x^{\alpha-1}(1-x)^{\beta-1} $
Then $\displaystyle E\left(\frac{2}{1+x}\right)=\int_{-\infty}^\infty \frac{2}{1+x}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\, x^{\alpha-1}(1-x)^{\beta-1}dx$
$\displaystyle E\left(\frac{2}{1+x}\right)=\frac{\Gamma(3/2)}{\Gamma(2)\Gamma(1/2)}\int_{-\infty}^\infty \frac{2}{1+x}\, x^{1}(1-x)^{-1/2}dx$
Now what do I do ? Please help.