7

How do I show that

Let $n\ge0$. Then, $$I:=\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!}$$ Here, $m!!$ denotes the product of all positive integers $i \in \left\{1,2,\ldots,m\right\}$ that have the same parity as $m$.

My try: Using Binomial theorem

$$(1-x^2)=1-nx^2+{n(n+1)\over2!}x^4-{n(n+1)(n+2)\over 3!}x^6+\cdots$$

$$\int_{0}^{1}\left(1-nx^2+{n(n+1)\over2!}x^4-{n(n+1)(n+2)\over 3!}x^6+\cdots\right)dx$$

$$I=x-n{x^3\over 3}+{n(n+1)\over 2!}{x^5\over 5}-{n(n+1)(n+2)\over 3!}{x^7\over 7}+\cdots|_{0}^{1}$$

$$I=1-{n\over 1\cdot 3}+{n(n+1)\over 2!\cdot 5}-{n(n+1)(n+2)\over 3!\cdot 7}+\cdots$$

I need help can't see how it will simplify to ${(2n)!!\over (2n+1)!!}$

darij grinberg
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gymbvghjkgkjkhgfkl
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2 Answers2

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Hint. Integrating by parts gives $$ \begin{align} I_n=\int_{0}^{1}(1-x^2)^ndx&=\left[x(1-x^2)^n\right]_{0}^{1}+2n\int_{0}^{1}x^2(1-x^2)^{n-1}dx \\\\&=0+2n\int_{0}^{1}\left[(1-(1-x^2))(1-x^2)^{n-1}\right]dx \\\\&=2nI_{n-1}-2nI_{n} \end{align} $$ then, with $I_0=1,\,I_1=\frac23,$ $$ I_{n}=\frac{2n}{2n+1}\cdot I_{n-1}, \quad n\ge1. $$ I think you can take it from here.

Olivier Oloa
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2

Let $z=x^{2}$ \begin{align} \int\limits_{0}^{1} (1-x^{2})^{n} dx &= \frac{1}{2} \int\limits_{0}^{1} (1-z)^{n} z^{-1/2} dz \\ &= \frac{1}{2} \mathrm{B}(1/2,n+1) \\ &= \frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma(n+3/2)} \\ \tag{1} &= \frac{\sqrt{\pi}n!}{(2n+1)\Gamma(n+1/2)} \\ \tag{2} &= \frac{\sqrt{\pi}n!}{(2n+1)} \frac{2^{n}}{\sqrt{\pi}(2n-1)!!} \\ &= \frac{n!2^{n}}{(2n+1)} \frac{n!2^{n}}{(2n)!} \\ &= \frac{(2n)!!n!2^{n}}{(2n+1)!} \\ &= \frac{(2n)!!n!2^{n}}{(2n+1)!!n!2^{n}} \\ &= \frac{(2n)!!}{(2n+1)!!} \end{align}

  1. $\Gamma(n+3/2) = \Gamma((n+1/2)+1) = (n+1/2)\Gamma(n+1/2)$

  2. $$\Gamma(n+1/2) = \frac{(2n-1)!!\sqrt{\pi}}{2^{n}}$$

  3. All double factorial identities used can be found here.

poweierstrass
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