Method for solving $\int_0^1 \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^\alpha (1-x)^\beta dx$ I really don't know where to start in integrating this
$\int_0^1 \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^\alpha (1-x)^\beta dx$
 A: The fraction containing the gamma function is called the beta function. Now I assume that $\alpha$ and $\beta$ are constants.
The beta function is
\begin{equation}
B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
\end{equation}
So for your integral
\begin{align}
\int_0^1 \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} x^\alpha (1-x)^\beta dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} \int_0^1 x^\alpha (1-x)^\beta dx
\end{align}
Now the integral is just the beta function $B(\alpha + 1, \beta + 1)$, so
\begin{align}
\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} \int_0^1 x^\alpha (1-x)^\beta dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} \cdot \frac{\Gamma(\alpha + 1)\Gamma(\beta + 1)}{\Gamma(\alpha + \beta + 2)}
\end{align}
You can now simplify from this point using the identity $\Gamma(x + 1) = x\Gamma(x)$.
A: Hint: Consider 
$$ g(\alpha,\beta)=\int_{0}^{1}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}\,dx $$
for $\alpha,\beta\in\mathbb{Z}^+$ and prove by integration by parts and induction (on $\alpha$, then on $\beta$) that $g(\alpha,\beta)$ constantly equals $1$ on $\mathbb{Z}^+\times\mathbb{Z}^+$. Use the properties of the $\Gamma$ function (its log-convexity, for instance) to prove that the same holds for $\alpha,\beta > 0$ and deduce that:
$$ \int_{0}^{1}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha}(1-x)^{\beta}\,dx  =\frac{\alpha \beta}{(\alpha+\beta+1)(\alpha+\beta)}.$$
A: Hint you can always start by reading your courses, more precisely the linearity of integration and the definition of beta function or the gamma function ...
