Integral of a binomial-like function through Integration by parts I wonder how to calculate the following through integration by parts:
\begin{align*}
\int_{0}^{1}\binom{n}{i}u^{i}(1-u)^{n-i}\, du
\end{align*}
$i=0,1,2,\cdots,n$. I suppose Gamma function might be useful here?
 A: Let
$$I:=\int_0^1 \binom{n}{i} u^i(1-u)^{n-i} \text{d}u$$
So
$$(n+1)I=\sum_{i=0}^{n}\int_0^1 \binom{n}{i} u^i(1-u)^{n-i} \text{d}u=\int_0^1 \sum_{i=0}^{n}\binom{n}{i} u^i(1-u)^{n-i} \text{d}u$$
$$=\int_0^11^n\text{d}u=1\implies I=\frac{1}{n+1}$$
A: Using $\beta$-integral:
$$I={n \choose k} \int_{0}^{1}x^k(1-x)^{n-k} dx={n \choose k} \frac{\Gamma(k+1) \Gamma(n-k+1)}{\Gamma(n+2)}={n \choose k}\frac{k! (n-k)!}{(n+1)!}=\frac{1}{n+1}$$
A: Indeed integration by parts is quite straightforward: Writing $I_i$ for the given integral and setting $U = u^i$ with $dU = i u^{i-1} du$ and $dV = (1-u)^{n-i} du$ with $V = -\frac{1}{n-i+1}(1-u)^{n-i+1}$ (where we assume $i\le n$), we have
$$
I_i = \binom{n}{i} \left[ \left.(\dots) u^i (1-u)^{n-i+1}\right|_0^1 + \frac{i}{n-i+1} \int_0^1 u^{i-1}(1-u)^{n-(i-1)} du \right] = I_{i-1}
$$
The first term is zero at both boundaries for $1 \le i \le n$, and the second term gives $I_{i-1}$, using
$$
\binom{n}{i}\cdot\frac{i}{n-i+1} = \frac{n!}{i!(n-i)!}\cdot\frac{i}{n-(i-1)} = \frac{n!}{(i-1)!(n-(i-1))!} = \binom{n}{i-1}
$$
