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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?

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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 $$ I_n=\frac{n}{4n+2}I_{n-1}\tag{5} $$

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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.

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

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  • $\begingroup$ 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). $\endgroup$ – robjohn Mar 18 '12 at 21:16
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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).

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  • $\begingroup$ 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 $x=sin^2(t)$. $\endgroup$ – Gabi Purcaru Mar 18 '12 at 20:18

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