Proving ${n \choose k}^{-1} = (n+1)\int_0^1 x^k(1-x)^{n-k}\mathsf dx$

Question

Title says it all, I've tried to find the indefinite integral of the right side, got some sort of weird series and got stuck:

$$\sum_{i=0}^n-k {n-k \choose i}\cdot{(-1)i\over k+1+i}$$

Answer 1

It is easy to see using the definition that $$ \binom{n}{k} = \frac{n}{k}\binom{n-1}{k-1} = \frac{n(n-1)}{k(k-1)}\binom{n-2}{k-2} = \dotsb $$ Since $0 \leqslant k \leqslant n$ for this to make sense, we can iterate this until we get down to $\binom{n-k}{0}=1$. Therefore we have to prove that the integral satisfies the same recurrence relation (upside-down) and boundary condition. This is easy to do with the right integration by parts: $$ \begin{align*} &(n+1)\int_0^1 x^k(1-x)^{n-k} \, dx \\ &= (n+1)\left[ -\frac{(1-x)^{n+1}}{n+1} \left( \frac{x}{1-x} \right)^k \right]_0^1 + \int_0^1 (1-x)^{n+1} \left( k x^{k-1}(1-x)^{-k} + kx^k(1-x)^{-k-1} \right) \, dx \\ &= 0 - k \int_0^1 x^{k-1}(1-x)^{n+1} (1-x)^{-k-1} (1-x+x) \, dx \\ &= \frac{k}{n} \left( n \int_0^1 x^{k-1} (1-x)^{n-k} \, dx \right) \end{align*}, $$ so the integral satisfies the same recurrence as $\binom{n}{k}^{-1}$. All we have to do now is check the boundary conditon, which is just $$ (n+1)\int_0^1 x^{k-k} (1-x)^{n-k+k} \, dx = 1, $$ so the result follows.

It's one of those lovely results where you choose the right integration by parts to start with, and then everything just works. And even better, you don't ever need $n$ to be an integer.