# Thread: Problem #21: Recurrence Integral

1. ## Problem #21: Recurrence Integral

Show that
$\displaystyle \int_0^1 (1 - x^2)^n~dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$

Use any technique you like. It looks, perhaps, like something using Legendre polynomials? (I didn't do it this way)

One way to attack this (and the one I'll be featuring as my solution) is to denote the integral as $\displaystyle I_n$. We have the recurrence relation:
$\displaystyle I_{k + 1} = \left ( \frac{2k + 1}{2k + 3} \right )~ I_k$

-Dan

2. ## Re: Problem #21: Recurrence Integral

$\displaystyle \int_0^1 (1-x)^n \, dx = \frac{1}{n+1}$

3. ## Re: Problem #21: Recurrence Integral

Originally Posted by Idea
$\displaystyle \int_0^1 (1-x)^n \, dx = \frac{1}{n+1}$
Whoops! Typo. I have fixed it in the problem statement. Thanks for the catch!

-Dan

4. ## Re: Problem #21: Recurrence Integral

$\displaystyle I_{k+1}=\frac{2 (k+1)}{2 k+3}I_k$

5. ## Re: Problem #21: Recurrence Integral

That's the last time I post a problem on two hours of sleep. Thanks (again) for the catch.

Originally Posted by topsquark
Show that
$\displaystyle \int_0^1 (1 - x^2)^n~dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$

Use any technique you like. It looks, perhaps, like something using Legendre polynomials? (I didn't do it this way)

One way to attack this (and the one I'll be featuring as my solution) is to denote the integral as $\displaystyle I_n$. We have the recurrence relation:
$\displaystyle I_{k + 1} = \left ( \frac{2k + 2}{2k + 3} \right )~ I_k$
-Dan

6. ## Re: Problem #21: Recurrence Integral

No takers this time around. Let's see if I can post the solution without any typos!

Spoiler:
Let's look at the recurrence relation first.

Let $\displaystyle I_{k + 1} = \int_0^1 (1 - x^2)^{k + 1}~dx$

Do this by parts:
$\displaystyle \int u ~ dv = uv - \int v~du$

I am choosing $\displaystyle u = (1 - x^2)^{k + 1}$ and $\displaystyle dv = dx$

This gives
$\displaystyle I_{k + 1} = \left . x \cdot (1 - x^2)^{k + 1} \right |_0^1 - \int_0^1 (k +1)x(1 - x^2)^k(-2x)~dx$

The first term vanishes leaving
$\displaystyle I_{k + 1} = (2k + 2) \int_0^1 x^2 (1 - x^2)^k~dx$

Now for a cool little trick:
$\displaystyle I_{k + 1} = (2k + 2) \int_0^1 \left [1 - (1 - x^2) \right ] (1 - x^2)^k~dx$

$\displaystyle I_{k + 1} = (2k + 2) \left [ \int_0^1 (1 - x^2)^k~dx - \int_0^1 (1 - x^2)^{k + 1}~dx \right ]$

$\displaystyle I_{k + 1} = (2k + 2)I_k - (2k + 2)I_{k + 1}$

So finally
$\displaystyle I_{k + 1} = \left ( \frac{2k + 2}{2k + 3} \right ) I_k$

Now for the induction step:

Base case: k = 0:
$\displaystyle \int_0^1 ( 1 - x^2)^0~dx = 1 = \frac{2^{2 \cdot 0} (0!)^2}{1!}$ (Check!)

Now let's assume the theorem is true for some integer k. We need to show that the k + 1 case works as well. So:

$\displaystyle I_{k +1} = \left ( \frac{2k + 2}{2k + 3} \right ) I_k = \left ( \frac{2k + 2}{2k + 3} \right ) \cdot \frac{2^{2k} ( k!)^2}{(2k + 1)!} = \frac{2(k + 1)2^{2k}(k!)^2}{(2k + 3)(2k + 1)!}$

Multiply by 1:
$\displaystyle I_{k + 1} = \frac{2(k + 1)}{2k + 2} \cdot \frac{2(k + 1)2^{2k}(k!)^2}{(2k + 3)(2k + 1)!}$

$\displaystyle I_{k + 1} = \frac{ \left [ 2(k + 1) \right ] ^2 2^{2k}(k!)^2}{(2k + 3)(2k + 2)(2k + 1)!}$

$\displaystyle I_{k + 1} = \frac{2^{2(k + 1)} \left [ (k + 1)! \right ]^2}{\left [ 2(k + 1) + 1 \right ]!}$

So we have that the kth case implies the k+1th case. Therefore etc.

-Dan