Integrating the joint posterior function $\int_{0}^{\infty} k p^2 (1-p)^2 p^{\sum_{i=1}^{n}} x_{i} (1-p)^{n-\sum_{i=1}^{\infty}x_{i}} dp$

Question

So I'm trying to integrate this

$$\displaystyle \int_{0}^{\infty} k p^2 (1-p)^2 p^{\sum_{i=1}^{n} x_{i}} (1-p)^{n-\sum_{i=1}^{\infty}x_{i}} dp$$ where $k = \frac{\gamma(6)}{\gamma(3)\gamma(3)}$

Can anyone help me?

Answer 1

Use the formula for Beta function $$B(x,y) = \int_0^\infty t^{x-1} (1-t)^{y-1} dt$$ with $x = 3+\sum_i x_i$ and $y = 3 + n - \sum_i x_i$, so that $$\begin{aligned} & \int_{0}^{\infty} k p^2 (1-p)^2 p^{\sum_{i=1}^{n} x_{i}} (1-p)^{n-\sum_{i=1}^{\infty}x_{i}} dp \\ =& \frac{\gamma(6)}{\gamma(3)\gamma(3)} B\left( 3+\sum_{i=1}^n x_i,3 + n - \sum_{i=1}^n x_i \right). \end{aligned}$$