Binomial expansion of $\sqrt{1-x}$

Question

Show that $$\int_0^1x^{n-2}\sqrt{x(1-x)}dx=-\pi c_n, n=2,3,...$$ where $c_n$ is the coefficient of $x_n$ in the binomial expansion of $\sqrt{1-x},|x|<1$

$ (1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ... $

$c_n=\prod_{i=0}^{n-1}(\frac{1}{2}-i)/{n!}$

I guess I will have to do a contour integral to calculate the left-hand side.

Answer 1

Using the Beta and $\Gamma$ functions: $$ \int_{0}^{1}x^{n-2}\sqrt{x(1-x)}\,dx = \int_{0}^{1}x^{n-3/2}(1-x)^{1/2}\,dx =B\left(n-\tfrac{1}{2},\tfrac{3}{2}\right)=\frac{\Gamma\left(n-\tfrac{1}{2}\right)\,\Gamma\left(\tfrac{3}{2}\right)}{\Gamma\left(n+1\right)}$$ so the LHS equals $\frac{1}{n!}\,\Gamma\left(n-\tfrac{1}{2}\right)\Gamma\left(\tfrac{3}{2}\right)$. On the other hand $\Gamma\left(n-\tfrac{1}{2}\right)=\left(n-\tfrac{3}{2}\right)\,\Gamma\left(n-\tfrac{3}{2}\right)$, hence: $$ \Gamma\left(n-\tfrac{1}{2}\right)\Gamma\left(\tfrac{3}{2}\right)=\left(n-\tfrac{3}{2}\right)\,\Gamma\left(n-\tfrac{3}{2}\right)\Gamma\left(\tfrac{3}{2}\right)=\Gamma\left(\tfrac{3}{2}\right)\Gamma\left(-\tfrac{1}{2}\right)\prod_{k=1}^{n}\left(n-\tfrac{2k+1}{2}\right) $$ where the RHS equals (by setting $k=n-i$) $$ -\pi\prod_{i=0}^{n-1}\left(\frac{1}{2}-j\right)$$ as wanted.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}x^{n - 2}\root{x\pars{1-x}}\dd x & = \int_{0}^{1}x^{n - 3/2}\pars{1 - x}^{1/2}\,\dd x = \bbox[10px,#ffd]{\ds{{% \Gamma\pars{n - 1/2}\Gamma\pars{3/2} \over \Gamma\pars{n - 1}}}} \label{1}\tag{1} \\[1cm] c_{n} & \equiv \bracks{x^{n}}\root{1 - x} = {1/2 \choose n}\pars{-1}^{n} = {\pars{1/2}! \over n!\pars{1/2 - n}!}\,\pars{-1}^{n} \\[5mm] & = {\Gamma\pars{3/2} \over \Gamma\pars{n + 1}\Gamma\pars{3/2 - n}}\,\pars{-1}^{n} \\[5mm] & = {\Gamma\pars{3/2} \over \Gamma\pars{n + 1}\pars{\pi/\braces{\Gamma\pars{n - 1/2}\sin\pars{\pi\bracks{n - 1/2}}}}}\,\pars{-1}^{n} \\[5mm] & = {1 \over \pi}\, \bbox[#ffd,10px]{\ds{% {\Gamma\pars{n - 1/2}\Gamma\pars{3/2} \over \Gamma\pars{n - 1}}}}\, {1 \over n\pars{n - 1}}\label{2}\tag{2} \\[1cm] & \stackrel{\eqref{1}\ \mbox{and}\ \eqref{2}}{\implies}\,\,\, \bbx{\int_{0}^{1}x^{n - 2}\root{x\pars{1-x}}\dd x = \pi n\pars{n - 1}c_{n}} \end{align}

Answer 3

As @Jack mentions in the comments, your integral can be evaluated using the Beta and Gamma functions.

Here \begin{align*} \int_0^1 x^{n - 2} \sqrt{x(1 - x)} \, dx &= \int_0^1 x^{(n - \frac{1}{2}) - 1} (1 - x)^{\frac{3}{2} - 1} \, dx\\ &= \text{B} \left (n - \frac{1}{2}, \frac{3}{2} \right )\\ &= \frac{\Gamma \left (n - \frac{1}{2} \right ) \Gamma \left (\frac{3}{2} \right )}{\Gamma (n + 1)}\\ &= \frac{\sqrt{\pi}}{2 n!} \Gamma \left (n - \frac{1}{2} \right ), \end{align*} where we have made use of the facts that $\Gamma (\frac{1}{2}) = \sqrt{\pi}$ and $\Gamma (n + 1) = n!$.

Now, as $$\Gamma \left (n + \frac{1}{2} \right ) = \left (n - \frac{1}{2} \right ) \Gamma \left (n - \frac{1}{2} \right ),$$ then $$\Gamma \left (n - \frac{1}{2} \right ) = \frac{1}{(n - 1/2)} \Gamma \left (n + \frac{1}{2} \right ).$$ Now it can be readily shown that $$\Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)! \sqrt{\pi}}{2^{2n} n!}.$$ Thus $$\Gamma \left (n - \frac{1}{2} \right ) = \frac{1}{(n - 1/2)} \frac{(2n)! \sqrt{\pi}}{2^{2n} n!},$$ and the integral becomes $$\int_0^1 x^{n - 2} \sqrt{x(1 - x)} \, dx = \frac{\pi}{(2n - 1) 2^{2n}} \frac{(2n)!}{(n!)^2} = \frac{\pi}{(2n - 1) 2^{2n}} \binom{2n}{n} = -\pi c_n,$$ and implies that $$c_n = -\frac{1}{2^{2n}(2n - 1)} \binom{2n}{n}, \quad n = 2,3,4,\ldots$$

Now, from the binomial expansion for $\sqrt{1- x}, |x| < 1$, we have \begin{align*} \sqrt{1 - x} &= \sum_{n = 0}^\infty (-1)^n \binom{\frac{1}{2}}{n} x^n\\ &= \sum_{n = 0}^\infty (-1)^n \frac{\frac{1}{2}!}{(\frac{1}{2} - n)! n!}\\ &= \sum_{n = 0}^\infty (-1)^n \frac{\sqrt{\pi}}{2 \Gamma \left (\frac{3}{2} - n \right ) n!}\\ &= -\sum_{n = 0}^\infty \frac{1}{(2n - 1) 2^{2n}} \binom{2n}{n} x^n\\ &= \sum_{n = 0}^\infty c_n x^n, \end{align*} where $$c_n = -\frac{1}{2^{2n}(2n - 1)} \binom{2n}{n},$$ since $$\Gamma \left (\frac{3}{2} - n \right ) = \frac{(-1)^{n - 1} 2^{2n - 1} (2n - 1) n!}{(2n)!} \sqrt{\pi}. \tag1$$

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Addendum

Here is a proof of (1). To prove the result we will make use of the following result for the Gamma function $$\Gamma (x + 1) = x \Gamma (x) \quad \Rightarrow \quad \Gamma (x) = \frac{\Gamma (x + 1)}{x}.$$ Repeated application of the above result we see that \begin{align*} \Gamma \left (\frac{3}{2} - n \right ) &= \frac{\Gamma \left (\frac{2n + 1}{2} - n \right )}{\underbrace{\left (\frac{3}{2} - n \right ) \left (\frac{5}{2} - n \right ) \cdots \left (\frac{2n - 1}{2} - n \right )}_{(n - 1) \,\, \text{terms}}}\\ &= \frac{\Gamma \left (\frac{1}{2} \right ) 2^{n - 1}}{(-1)^{n - 1} (2n - 3) (2n - 5) \cdots 3 \cdot 1}\\ &= \frac{(-1)^{n - 1} 2^{n - 1} \overbrace{(2n - 2)(2n - 4) \cdots 2}^{(n - 1) \, \, \text{terms}}}{\underbrace{(2n - 2)(2n - 3)(2n - 4) \cdots 2 \cdot 1}_{(2n - 2) \,\, \text{terms}}} \sqrt{\pi}\\ &= \frac{(-1)^{n - 1} 2^{n - 1} \cdot 2^{n - 1} (n - 1)!}{(2n - 2)!} \sqrt{\pi}\\ &= \frac{(-1)^{n - 1} 2^{2n - 2} 2n (2n - 1) n!}{n (2n)!} \sqrt{\pi}\\ &= \frac{(-1)^{n - 1} 2^{2n - 1} (2n - 1) n!}{(2n)!} \sqrt{\pi}, \end{align*} as required to prove.