Integral with binomial coefficient Is it possible to evaluate this integral without using the gamma function
$$ \int_0^1 {a \choose b}x^b(1-x)^{a-b} dx$$
It looks a little like part of binomial theorem, but I don't have an idea how to manage with that. The solution should be $\frac{b}{a+1}$.
 A: Let $$I(a,b) = \binom{a}{b} \int_{x=0}^1 x^b (1-x)^{a-b} \, dx.$$ Integration by parts approach, with the choice $u = x^b$, $du = bx^{b-1}$, $dv = (1-x)^{a-b} \, dv$, $\displaystyle v = -\frac{(1-x)^{a-b+1}}{a-b+1}$, gives $$\begin{align*} I(a,b) &= \binom{a}{b} \left( \biggl[ -x^b \frac{(1-x)^{a-b+1}}{a-b+1} \biggr]_{x=0}^1 + \frac{b}{a-b+1} \int_{x=0}^1 x^{b-1} (1-x)^{a-b+1} \, dx \right) \\ &= \binom{a}{b}\frac{b}{a-b+1} \binom{a}{b-1}^{-1} I(a,b-1) \\ &= I(a,b-1). \end{align*}$$  By recursion, if $b$ is a positive integer, we then have $$I(a,b) = I(a,0) = \frac{1}{a+1}$$ by direct computation.  More sophisticated methods of evaluation can show that this result holds true as long as $-1 < b < a+1$ and $a \ne -1$ but since you do not want a treatment with the gamma function, it is a bit tedious to speak of the meaning of binomial coefficients of non-integer values.
A: Consider the generating function $F(y) = \sum_{b=0}^a  y^b \int_0^1 \binom{a}{b} x^b (1-x)^{a-b} \mathrm{d}x$. By the binomial theorem we have  
$$F(y) = \int_0^1 (1-(1-y)x)^a \mathrm{d} x = \frac{1}{a+1} \frac{1-y^{a+1}}{1-y} = \frac{1}{a+1}(1+y+\cdots+y^a) $$
Now take the coefficient of $y^b$ to get $\int_0^1 \binom{a}{b} x^b (1-x)^{a-b} \mathrm{d}x = 1/(a+1)$.
A: I assume $a > b$, and $a, b \in \mathbb{N}$. This is the integral of the kernel of a Beta density with parameters $\alpha = b+1$ and $\beta = a-b+1$, so use the fact that
\begin{align}
   1
&= \frac{1}{B(\alpha,\beta)}\int_0^1 x^b (1-x)^{a-b}dx \\
&= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1 x^b (1-x)^{a-b}dx \\
&= \frac{(a+1)!}{b!(a-b)!}\int_0^1 x^b (1-x)^{a-b}dx \\
&= (a+1)\int_0^1 \binom{a}{b}x^b (1-x)^{a-b}dx \\
\end{align}
EDIT
Here is another approach. Write the integral (without the binomial coefficient) as
\begin{align}
   \int_0^1 x^b(1-x)^{a-b}dx
&= \int_0^1 x^b\sum_{i=0}^{a-b} \binom{a-b}{i} (-x)^i dx \\
&= \sum_{i=0}^{a-b} \binom{a-b}{i}(-1)^i \int_0^1 x^{b+i} dx \\
&= \sum_{i=0}^{a-b} \binom{a-b}{i}(-1)^i \frac{1}{b+i+1} ,
\end{align}
and now multiply both sides by $\binom{a}{b}$. There may be some useful binomial identities to go further, but are you sure about the answer you gave?
