Proving that $\int_{0}^{1}x^{n-1}(1-x)^n \mathrm dx =\frac{1}{n {2n\choose n}}$

Question

I am working on the following problem and I've arrived at an integral which I need to show equals the proposition.

Prove that $\displaystyle \sum_{k=0}^{k=n}\dfrac{(-1)^{k}{n\choose k}}{n+k}=\dfrac{1}{n{2n\choose n}}$.

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My Attempt:

Let us consider for the sake of convenience the notation ${n\choose k}=\text{C}_k$ $$\begin{aligned}(1-x)^n&=\text{C}_0-\text{C}_1x+\text{C}_2x^2-\text{C}_3x^3+\ldots+(-1)^{n}\text{C}_nx^n\\ \int _{0}^{1}x^{n-1}(1-x)^n \mathrm dx&=\int_{0}^{1}\text{C}_0x^{n-1}-\text{C}_1x^n+\text{C}_2x^{n+1}+\ldots+(-1)^{n}\text{C}_nx^{2n-1}\mathrm dx\\ \int_{0}^{1}x^{n-1}(1-x)^n \mathrm dx&=\sum_{k=0}^{k=n}\dfrac{(-1)^{n}\text{C}_k}{n+k}\end{aligned}$$

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I now need to prove that the integral equals $\left[{2n\choose n}\right]^{-1}$. But I am unclear on how to proceed with simplifying the integral. I would appreciate any hints apart from usage of Integration by Parts technique. Thanks

Answer 1

Let's put $$I_{k,n} = \int_0^1 x^k (1-x)^n dx.$$ By integration by parts we have: $$I_{k,n} = \frac{n}{k+1} I_{k+1, n-1}. $$ Now we can iterate this to get $$ I_{k,n} = \frac{n}{k+1} I_{k+1,n-1} = \frac{n}{k+1} \frac{n-1}{k+2} I_{k+2, n-2} = \ldots = \frac{n!}{(k+1)(k+2) \ldots (k+n)} I_{k+n,0}. (*)$$

We have $$I_{k+n,0} = \int_0^1 x^{k+n} dx = \frac{1}{k+n+1}.$$ Now it remains to write the fraction from $(*)$ as a binomial coefficient.

Answer 2

$$I=\int_{0}^{1}x^{n-1}(1-x)^n dx= \frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)} =\frac{(n-1)! n!}{(2n)!}=\frac{1}{n{2n \choose n}}$$

Answer 3

$$\dfrac{d(x^m(1-x)^n)}{dx}=mx^{m-1}(1-x)^n-nx^m(1-x)^{n-1}$$

Integrate both sides wrt $x,$ $$x^m(1-x)^n=mI(m-1,n)-nI(m,n-1)$$ where $$I(m,n)=\int x^m(1-x)^n\ dx,J(m,n)=\int_0^1x^m(1-x)^n\ dx$$

and $$J(m-1,n)=\dfrac nm\cdot J(m,n-1)$$ $$=\dfrac{n(n-1)}{m(m-1)}\cdot J(m+1,n-2)$$ $$=\cdots$$ $$=\dfrac{n(n-1)\cdots(n-r+1)}{m(m-1)\cdots(m-r+1)}\cdot J(m+r-1,n-r)$$

$r=n\implies$ $$J(m-1,n)=\dfrac{n! }{m(m-1)\cdots(m-n+1)}\cdot J(m+n-1,0)$$

$\displaystyle J(m+n-1,0)=\dfrac1{m+n}$

Finally set $m=n$

Answer 4

Not sure if this helps,

Let $$I = \int_{0}^{1}x^{n-1}(1-x)^{n}dx = \int_{0}^{1}(1-x)^{n-1}x^{n}dx$$ Therefore $$2I = \int_{0}^{1}x^{n-1}(1-x)^{n-1}dx$$ Let, $$x = sin^2(u) $$ $$dx = 2sin(u)cos(u)du$$ $$2I = \int_{0}^{\pi/2}2sin^{2n-1}(u)cos^{2n-1}(u)du$$ for which standard results exist