-1

Let $$I(n)=\int_0^1 (x-x^2)^n dx .$$ Mainly, what I'm trying to get is a recurrent form of this integral that probably involves $I(n-1)$.

My attempt $(x-x^2)^n=(-(x^2-x+1/4-1/4))^n=(1/4-(x-1/2)^2)^n$. Let $t=x-\frac12$, hence the integral becomes

$$I(n)=\int_{1/2}^{3/2} \left(\frac14-t^2\right)^n dt .$$

Further , I tried integrating by parts , but I didn't manage to get anything .

Can someone try and find a reduction formula for this ?

Victor
  • 3,238
  • 1
    $ x^2 - x = -(x^2 - x) = - ((x-1/2)^2 - 1/4) = 1/4 - (x-1/2)^2 $

    Use substitution, $ y = x - 1/2 $ and $ y = 1/2 \cdot \sin \theta $, Wallis Product

     – GohP.iHan May 20 '14 at 19:59
  • 1
    $(x - x^2)^n = x^n(1 - x)^n$ should allow you to integrate by parts without any transformation. – Tunococ May 20 '14 at 19:59
  • @GohPiHan , oops , thanks a lot for spotting that stupid mistake :D – Victor May 20 '14 at 20:01
  • 1
    If you are not restricted to a recurrence that must only contain terms with $I(n)$, consider the beta function $B(p, q)=\int_{0}^{1}{x^{p-1}(1-x)^{q-1} dx}$ so $I(n)=B(n+1, n+1)$. You can easily show that $pB(p, q+1)=qB(p+1, q)$, so that $B(p, q)=\frac{(p-1)!(q-1)!}{(p+q-2)!}B(p+q-1, 1)=\frac{(p-1)!(q-1)!}{(p+q-1)!}$, so $I(n)=\frac{(n!)^2}{(2n+1)!}$. – Ivan Loh May 20 '14 at 20:10
  • I'm a 12th grade student , I guess I'm restricted to a recurrence that implies I(n) directly . Sorry for not mentioning this , I had no clue that this problem can have such various approaches. – Victor May 20 '14 at 20:12

3 Answers3

2

Well... I guess if you are not allowed to use the beta function, you can do as follows:

\begin{align} I(n)&=\int_{0}^{1}{x^n(1-x)^n dx} \\ &= \int_{0}^{1}{x^n(1-x)^n(x+(1-x))^2 dx} \\ &=\int_{0}^{1}{x^{n+2}(1-x)^n dx}+2\int_{0}^{1}{x^{n+1}(1-x)^{n+1} dx}+\int_{0}^{1}{x^n(1-x)^{n+2} dx} \\ &=2I(n+1)+2\int_{0}^{1}{x^{n+2}(1-x)^n dx} \\ &=2I(n+1)+2\left(\left[-\frac{x^{n+2}(1-x)^{n+1}}{n+1}\right]^1_0+\frac{n+2}{n+1}\int_{0}^{1}{x^{n+1}(1-x)^{n+1} dx}\right) \\ &=2I(n+1)+2\frac{n+2}{n+1}I(n+1) \\ &=\frac{4n+6}{n+1}I(n+1) \end{align}

where from the third to fourth line, we have used the change of variables $x \to 1-x$ for the third integral, and done integration by parts to get to the fifth line.

Ivan Loh
  • 17,115
2

Integrate by parts

$$I_n=\int_{0}^{1}{x^n(1-x)^n dx}=-\frac{1}{2} \int_{0}^{1}(1-2x)'{x^n(1-x)^n dx}=-\frac{1}{2} (1-2x)x^n(1-x)^n|_0^1+$$$$+\frac{1}{2}\int_{0}^{1}n(1-2x)^2{x^{n-1}(1-x)^{n-1} dx}=\frac{n}{2} \int_{0}^{1}[1-4(x-x^2)]{x^{n-1}(1-x)^{n-1} dx}=\frac{n}{2}I_{n-1}-2nI_n.$$ $$I_n=\frac{n}{2(2n+1)}I_{n-1}. $$

medicu
  • 4,592
1

With Tunococ's suggestion, we can apply Beta function.

GohP.iHan
  • 1,416