Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $I_n=\int_0^1(x-x^2)^ndx$. Prove that $I_n=\frac{1}{4}\cdot\frac{2n}{2n+1}I_{n-1}$.

This sounds like a rather easy exercise, but no matter how hard I try, I can't quite put my finger on it (I tried integration by parts with $\int_0^1x'(x-x^2)^ndx$, but couldn't manage to get too far). Could you help me out?

share|improve this question

3 Answers 3

up vote 11 down vote accepted

Integrating by parts in two different ways gives $$ \begin{array}{rllr} I_n &=\int_0^1x^n(1-x)^n\,\mathrm{d}x&=\hphantom{\frac{n}{n+1}}\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{x(1-x)}\,\mathrm{d}x&\qquad(1)\\ &=\frac{n}{n+1}\int_0^1x^{n+1}(1-x)^{n-1}\,\mathrm{d}x&=\frac{n}{n+1}\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{x^2}\,\mathrm{d}x&(2)\\ &=\frac{n}{n+1}\int_0^1x^{n-1}(1-x)^{n+1}\,\mathrm{d}x&=\frac{n}{n+1}\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{(1-x)^2}\,\mathrm{d}x&(3)\\ \end{array} $$ Since $2\color{red}{x(1-x)}+\color{red}{x^2}+\color{red}{(1-x)^2}=\color{red}{1}$, if we add $2(1)+\frac{n+1}{n}(2)+\frac{n+1}{n}(3)$, we get $$ 2I_n+\frac{n+1}{n}I_n+\frac{n+1}{n}I_n=\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{1}\,\mathrm{d}x=I_{n-1}\tag{4} $$ Thus, solving $(4)$ for $I_n$ yields $$ I_n=\frac{n}{4n+2}I_{n-1}\tag{5} $$

share|improve this answer
    
great one. Thanks! –  Gabi Purcaru Mar 18 '12 at 20:45

The integral $I_n$ is beta function evaluated at $(n+1, n+1)$:

$I_n=\int_0^1(x-x^2)^ndx = \int_0^1x^n(1-x)^ndx = B(n+1, n+1)$

Using the fact that $B(z,w) = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}$, where $\Gamma$ is the gamma function, the result follows.

$I_n = B(n+1, n+1) = \frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)} = \frac{n^2\Gamma(n)\Gamma(n)}{(2n+1)(2n)\Gamma(2n)} = \frac{1}{4}\cdot\frac{2n}{2n+1} B(n,n)$

share|improve this answer
    
This approach was my first thought. I went with a more basic method (essentially the way that the Beta relation is shown). This is a solid, if more advanced, approach (+1). –  robjohn Mar 18 '12 at 21:16

This is perhaps not the most direct way. Let $x = \sin^2(\theta)$. We then have $dx = 2 \sin(\theta) \cos(\theta) d \theta$. Hence, $$I_n = \int_0^{\pi/2} \sin^{2n}(\theta) \cos^{2n}(\theta) 2 \sin(\theta) \cos(\theta) d \theta$$ $$I_n = \frac1{2^{2n}} \int_0^{\pi/2} \sin^{2n+1}(2 \theta) d \theta = \frac1{2^{2n+1}} \int_0^{\pi} \sin^{2n+1}(\phi) d \phi = \frac1{2^{2n}} \int_0^{\pi/2} \sin^{2n+1}(\phi) d \phi$$ Let us denote $$J_{n} = \int_0^{\pi/2} \sin^{n}(\phi) d \phi.$$ Refer here for the recurrence involving $J_n$. We have $J_{2n+1} = \frac{2n}{2n+1} J_{2n-1}$. Hence, $$I_n = \frac{J_{2n+1}}{2^{2n}} = \frac14 \frac{2n}{2n+1} \frac{J_{2n-1}}{2^{2n-2}} = \frac14 \frac{2n}{2n+1} I_{n-1}.$$

EDIT: The integral $I_n$ is $\beta(n+1,n+1)$ where $\beta(x,y)$ is the $\beta$ function defined as $\displaystyle \beta(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$. $\beta(x,y)$ also has a representation in terms of the $\Gamma$ function given by $\displaystyle \beta(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$.

Hence, $$\begin{align} I_n & = \beta(n+1,n+1) = \frac{\Gamma(n+1) \Gamma(n+1)}{\Gamma(2n+2)} = \frac{n! n!}{(2n+1)!} = \frac{n^2}{(2n+1)(2n)}\frac{(n-1)! (n-1)!}{(2n-1)!}\\ & = \frac{n}{2(2n+1)} \frac{\Gamma(n) \Gamma(n)}{\Gamma(2n)} = \frac14 \frac{2n}{2n+1} \beta(n,n) = \frac14 \frac{2n}{2n+1} I_{n-1}. \end{align}$$

(I was adding this part when m. k. posted the answer).

share|improve this answer
    
thanks for the quick answer. I am hoping for a more straightforward solution if one exists, because this was proposed for the Romanian SAT-equivalent exam, so it should not make use of speculations like $x=sin^2(t)$. –  Gabi Purcaru Mar 18 '12 at 20:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.