# Beta function in Philip J. Davis׳ Essay

This question is about equation number (4) in Philip J. Davis’ Essay titled "LEONHARD EULER'S INTEGRAL: A HISTORICAL PROFILE OF THE GAMMA FUNCTION".

In there it is stated by the author "Euler expanded $$(1-x)^n$$ by the binomial theorem, and without difficulty found that:"

$$\int_0^1 x^e (1-x)^n dx = \frac{1·2···n}{(e+1)(e+2)···(e+n+1)}$$

In this case $$e$$ is an arbitrary value and $$n$$ is an integer. I tried to reproduce this by myself but I get stuck with the coefficients of the binomial, I do not see how to cancel them nicely to arrive at this solution by Euler.

Could somebody help out understanding Euler's work by providing a sketch on how to arrive to this equality?

• It would be easier to integrate by parts $n$ times. – user10354138 May 22 at 8:52
• Use induction on $n$. – Start wearing purple May 22 at 17:27
• Thanks for your comments, I indeed tried integration by parts, deriving (1-x)^n and integrating x^e, and it is straightforward to arrive to the equality. – jto May 23 at 9:51

\begin{align} \int_0^1 x^e (1-x)^n dx &= \int_0^1 x^e \sum_{k=0}^n \binom{n}{k} (-x)^kdx = \\ &= \sum_{k=0}^n \binom{n}{k} (-1)^k \int_0^1 x^{e+k} dx = \\ &= \sum_{k=0}^n \frac{n!}{k! (n-k)!} \frac{(-1)^k}{e+k+1} = \\ &= \frac{p(e)}{(e+1)(e+2)\dots(e+n+1)}\end{align} where $$p(e)$$ is a certain polynomial of degree $$n$$, specificaly $$p(e) = \sum_{k=0}^n \frac{n!}{k! (n-k)!} (-1)^k \prod_{i\in\{0,\dots,n\}\setminus\{k\}} (e+i+1)$$ If we put $$e=-j-1$$, $$j\in\{0,\dots, n\}$$ only one term in the sum remains and we get \begin{align} p(-j-1) &= \frac{n!}{j! (n-j)!} (-1)^j \prod_{i\in\{0,\dots,n\}\setminus\{j\}} (i-j) = \\ &= \frac{n!}{j! (n-j)!} (-1)^j \cdot (-1)^j j!(n-j)! = \\ &= n!\end{align} Since $$p(e)$$ takes value $$n!$$ in $$n+1$$ different points, and $$p(e)$$ is a polynomial of degree $$n$$, that means that $$p(e) =n!$$. Thus we get $$\int_0^1 x^e (1-x)^n dx = \frac{n!}{(e+1)(e+2)\dots(e+n+1)}$$