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Let $(I_n)_{n \geq 1}$ be a sequence such that: $$I_n = \int_0^1 (1-x^2)^n dx$$

Find the value of $I_n$ (The solution is $\frac{2}{3} \cdot \frac{4}{5} \cdot ... \cdot \frac{2n}{2n+1}$).

I've tried using the binomial expansion of $(1-x^2)^n$ but I couldn't get the proper answer.

Thank you!

Olivier Oloa
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George R.
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3 Answers3

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I've tried using the binomial expansion of $(1-x^2)^n$ but I couldn't get the proper answer.

One may use integration by parts obtaining $$ \begin{align} I_n=\int_{0}^{1}(1-x^2)^ndx&=\left[x(1-x^2)^n\right]_{0}^{1}+2n\int_{0}^{1}x^2(1-x^2)^{n-1}dx \\\\&=0+2n\int_{0}^{1}\left[(1-(1-x^2))(1-x^2)^{n-1}\right]dx \\\\&=2nI_{n-1}-2nI_{n} \end{align} $$ then, using $I_0=1$, $I_1=\frac23,$ $$ I_{n}=\frac{2n}{2n+1}\cdot I_{n-1}, \quad n\ge1, $$ one gets the announced result:

$$I=\int_{0}^{1}(1-x^2)^ndx=\frac{2}{3} \cdot \frac{4}{5} \cdots \frac{2n}{2n+1}={(2n)!!\over (2n+1)!!}, \qquad n\ge1.$$

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

$$=(1-x^m)^n-m(1-x^m)^{n-1}(1-(1-x^m))$$

$$\implies\dfrac{d(x(1-x^m)^n)}{dx}=(1-x^m)^n(1+m)-mn(1-x^m)^{n-1}$$

Integrate both sides wrt $x,$

$$x(1-x^m)^n=(mn+1)I_n-mnI_{n-1}$$ where $$I_n=\int(1-x^m)^n\ dx$$

lab bhattacharjee
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Here is an approach that relies on the Beta function $\operatorname{B}(m,n)$, namely $$\operatorname{B} (m,n) = \int_0^1 x^{m - 1} (1 - x)^{m - 1} \, dx.$$

In the integral for $I_n$ we begin by enforcing a substitution of $x \mapsto \sqrt{x}$. Thus \begin{align} I_n &= \frac{1}{2} \int_0^1 x^{-1/2} (1 - x)^n \, dx\\ &= \frac{1}{2} \int_0^1 x^{\frac{1}{2} - 1} (1 - x)^{(n + 1) - 1} \, dx\\ &= \frac{1}{2} \operatorname{B} \left (\frac{1}{2}, n + 1 \right )\\ &= \frac{1}{2} \frac{\Gamma (\frac{1}{2}) \Gamma (n + 1)}{\Gamma (n + \frac{3}{2} )}\\ &= \frac{\sqrt{\pi}}{2} \frac{\Gamma (n + 1)}{(n + \frac{1}{2}) \Gamma (n + \frac{1}{2})}. \end{align} Since $n$ is a positive integer, we have $$\Gamma (n + 1) = n! \quad \text{and} \quad \Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)!}{2^{2n} n!} \sqrt{\pi},$$ then $$I_n = \frac{2^{2n}}{2n + 1} \frac{(n!)^2}{(2n)!} = \frac{2^{2n}}{(2n + 1) \binom{2n}{n}}.$$ Here $\binom{2n}{n}$ denotes the central binomial coefficient.

omegadot
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