Expected value for $f(x)= \frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{\beta^\alpha}{\sqrt{\pi}} \frac{x^{\alpha-1}}{\sqrt{1-\beta x}}$ $$f(x)= \frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{\beta^\alpha}{\sqrt{\pi}} \frac{x^{\alpha-1}}{\sqrt{1-\beta x}}$$ 
where $0<x<\beta$.
So these are three terms all multiplied to give you an ugly distribution function where $\alpha>0$ is some parameter, $\beta>0$ is a parameter. $\Gamma$ refers to the Gamma function. 
This very closely resembles the Gamma Distribution function but not quite and I don't know how to find the expectation and variance for $X$ with the given distribution function.
I tried to go the route of finding the moment generating function to make the distribution resemble a gamma and use the fact that the density would integrate to one but the $(1-\beta x)$ term really complicates things. Not sure what to do. 
Help. 
 A: The expected value of a  probability density function $f(x)$ is given by
$$ \operatorname{E}[X] = \int_{-\infty}^\infty x f(x)\, \operatorname{d}x .$$
Applying this to your problem, we have
$$ E[X] = \frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{\beta^\alpha}{\sqrt{\pi}}\int_{0}^\beta   \frac{x^{\alpha}}{\sqrt{1-\beta x}}\, \operatorname{d}x  $$
$$ E[X] =  \frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{\beta^\alpha}{\sqrt{\pi}}\int_{0}^\beta x^{\alpha}(1-\beta x)^{-\frac{1}{2}}\, \operatorname{d}x . $$
Make the change of variables $y=\beta x$ yields
$$ E[X] = \frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{\beta^\alpha}{\sqrt{\pi}}\int_{0}^1 \frac{y^{\alpha}}{{\beta}^{\alpha}}(1- y)^{-\frac{1}{2}}\, \frac{dx}{\beta}$$
$$=\frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{\beta^\alpha}{\sqrt{\pi}}\int_{0}^1 \frac{y^{\alpha}}{{\beta}^{\alpha}}(1- y)^{-\frac{1}{2}}\, \frac{dx}{\beta} $$
$$  = \frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{1}{\beta\sqrt{\pi}} \int_{0}^1 y^{\alpha}(1- y)^{-\frac{1}{2}}\, \frac{dx}{\beta}$$
$$ = \frac{\Gamma (\alpha+\frac{1}{2})}{\Gamma (\alpha)} \frac{1}{\beta\sqrt{\pi}} \frac{\Gamma(\alpha+1)\Gamma(\frac{1}{2})}{\Gamma(\alpha + \frac{3}{2})}=\frac{\alpha}{\beta(\alpha+\frac{1}{2})}. $$
Note that, the last integral is known as the beta function
$$ \int_{0}^1 t^{u-1}(1- t)^{v-1}\, dt=\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}. $$ 
A: Let us take for granted that the function $f$ in the question, which we rewrite as $f_{\alpha,\beta}$, is a density function. In particular, $\displaystyle\int f_{\alpha,\beta}=1$ for every $\alpha$ and $\beta$. Now, for every $\gamma$,
$$
x^\gamma f_{\alpha,\beta}(x)=\frac{\Gamma(\alpha+\tfrac12)}{\Gamma(\alpha)}\,\frac{\Gamma(\alpha+\gamma)}{\Gamma(\alpha+\gamma+\tfrac12)}\,\frac1{\beta^\gamma}\,f_{\alpha+\gamma,\beta}(x).
$$
Since  $\displaystyle\int f_{\alpha+\gamma,\beta}=1$, this yields without any further computations that
$$
\mathbb E(X^\gamma)=\int x^\gamma f_{\alpha,\beta}(x)\mathrm dx=\frac{\Gamma(\alpha+\tfrac12)}{\Gamma(\alpha)}\,\frac{\Gamma(\alpha+\gamma)}{\Gamma(\alpha+\gamma+\tfrac12)}\,\frac1{\beta^\gamma}.
$$
Using this for $\gamma=1$ and $\gamma=2$, one gets
$$
\mathbb E(X)=\frac{\alpha}{(\alpha+\tfrac12)}\,\frac1\beta,\qquad\mathbb E(X^2)=\frac{\alpha(\alpha+1)}{(\alpha+\tfrac12)(\alpha+\tfrac32)}\,\frac1{\beta^2},
$$
from which the variance follows as
$$
\mathrm{var}(X)=\frac{\alpha}{2(\alpha+\tfrac12)^2(\alpha+\tfrac32)}\,\frac1{\beta^2}.
$$
Note that $X=Y/\beta$ where the distribution of $Y$ is Beta with parameters $(\alpha,\frac12)$, and an extended list of its properties is here.
A: Given: $X$ has pdf $f(x)$:

Notwithstanding lots of elaborate calculations by others on this page, unfortunately, the pdf itself is not well-defined. The easiest way to see this is to simply calculate the cdf $P(X<x)$ for some arbitrary parameter values ... I am using the mathStatica / Mathematica combo here:

which does not integrate to unity for parameter $\beta < 1$. For $\beta > 1$, it appears complex.  
A quick play suggests that it may be only well-defined for $\beta = 1$ ... in which case the distribution is just a special case of the Beta distribution, namely: $X$ ~ $Beta(\alpha, \frac 12)$.  

Addendum
In a commment below, 'Did' kindly confirms the above error  ... and notes that the flaw is obvious (which is perhaps why it has taken 5 months for anyone to notice it, user 'Did' included). With the suggested change, all is now well:

The desired mean and variance are now simply obtained as:

A: With a change of variables and the integral for the  Beta function, we get
$$
\begin{align}
\int_0^{1/\beta}\frac{x^{\alpha-1}}{\sqrt{1-\beta x}}\,\mathrm{d}x
&=\beta^{-\alpha}\int_0^1u^{\alpha-1}(1-u)^{-1/2}\,\mathrm{d}u\\
&=\beta^{-\alpha}\frac{\Gamma(\alpha)\sqrt\pi}{\Gamma(\alpha+1/2)}\tag{1}
\end{align}
$$
Thus,
$$
f(x)=\frac{\beta^\alpha}{\sqrt\pi}\frac{\Gamma(\alpha+1/2)}{\Gamma(\alpha)}\frac{x^{\alpha-1}}{\sqrt{1-\beta x}}\tag{2}
$$
has integral $1$. Using $(1)$, the expected value of $f(x)$ is
$$
\begin{align}
&\left.\int_0^{1/\beta}x\,f(x)\,\mathrm{d}x \middle/\int_0^{1/\beta}f(x)\,\mathrm{d}x\right.\\
&=\left.\int_0^{1/\beta}\frac{x^{\alpha}}{\sqrt{1-\beta x}}\,\mathrm{d}x \middle/\int_0^{1/\beta}\frac{x^{\alpha-1}}{\sqrt{1-\beta x}}\,\mathrm{d}x\right.\\
&=\frac{\alpha/\beta}{\alpha+1/2}\tag{3}
\end{align}
$$
