Beta function related integral Does anyone can help evaluating
$$\int^1_0xF(x)^{k}(1-F(x))^{n-k}f(x)dx$$ and 
$$\int^1_0x^2F(x)^{k}(1-F(x))^{n-k}f(x)dx?$$
Here, $F$ is a CDF with support $[0,1]$ and $f$ is its corresponding pdf. 
For the first one, substituting $F(x)=y$ yields
$$\int^1_0F^{-1}(y)y^{k}(1-y)^{n-k}dy.$$
Without the $F^{-1}$ term, it is a Beta function whose value is $\frac{k!(n-k)!}{(n+1)!}$, but with the $F^{-1}$ term, I don't know how to evaluate it or where to start doing it. Any help will be appreciated. 
 A: By the binomial development,
$$I=\int xF^k(x)\sum_{j=0}^{n-k}\binom{n-k}j(-1)^jF^{n-k-j}(x)f(x)\,dx\\
=\sum_{j=0}^{n-k}\binom{n-k}j(-1)^j\int xF^{n-j}(x)f(x)\,dx$$
Then by parts,
$$\sum_{j=0}^{n-k}\frac x{n-j+1}\binom{n-k+}j(-1)^jF^{n-j+1}(x)\,dx-\sum_{j=0}^{n-k}\frac1{n-j+1}\binom{n-k}j(-1)^j\int F^{n-j+1}(x)\,dx$$
and you are stuck.
A: Let's take a look at the first integral. First of all, you can use per partes method to get rid of $x$ term in the integral. Let me denote
$$
  A = \int_0^1 x F^k(x)(1-F(x))^{n-k}f(x)\,\mathrm{d}x.
$$
After per partes, we get
$$
  A
  =
  \left[
    x
    \int_0^x F^k(t)(1-F(t))^{n-k}f(t)\,\mathrm{d}t
  \right]_{x=0}^1
  -
  \int_0^1
  \int_0^x
  F^k(t)(1-F(t))^{n-k}f(t)\,\mathrm{d}t\,\mathrm{d}x.
$$
So we have
$$
  A = \int_0^1 F^k(t)(1-F(t))^{n-k}f(t)\,\mathrm{d}t
  -
  \int_0^1 \int_0^x F^k(t)(1-F(t))^{n-k}f(t)\,\mathrm{d}t\,\mathrm{d}x.
$$
Now we can use the substitution $y = F(t)$, $\mathrm{d}y = f(t)\,\mathrm{d}t$ and we get:
$$
  A = \underbrace{\int_0^1 y^k(1-y)^{n-k}\,\mathrm{d}y}_{B(k+1,n-k+1)}
  - \int_0^1 \underbrace{\int_0^{F(x)} y^k(1-y)^{n-k}\,\mathrm{d}t}_{B(F(x);k+1,n-k+1)}\,\mathrm{d}x
$$
where $B(x; a,b)$ is the incomplete Beta function [1]. I don't see how to continue now for general $F(x)$, but for $F(x) = x$ it is easy.

Update: I made an error. The following holds only if $F(x) = x$.
The integral of incomplete beta function can also be found in [1] while getting:
$$
  A
  =
  \frac{k!(n-k)!}{(n+1)!}
  -
  \left[
    x B(x;k+1,n-k+1) - B(x; k+2,n-k+1)
  \right]_{x=0}^1.
$$
It is easy to see that $B(0; a, b) = 0$ and we are left with
$$
  A
  =
  \frac{k!(n-k)!}{(n+1)!} - B(1;k+1,n-k+1) + B(1;k+2,n-k+1)
  \\
  = B(1;k+2,n-k+1) = B(k+2,n-k+1)
  \\
  = \frac{(k+1)!(n-k)!}{(n+2)!}.
$$

[1] http://mathworld.wolfram.com/IncompleteBetaFunction.html
