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We have the series $\left(I_n\right)_{n\ge 1\:}$ where $$I_n=\int _0^1\left(1-x^2\right)^n\,dx.$$

Prove that $$I_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.$$

I tried to integrate that function for $n=1,2,3$ to see whether there's a pattern (recurrence relation), but I just can't figure out how to write the pattern and how to prove that statement. Must I use induction?

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  • $\begingroup$ Induction and integration by parts is one way to go, but you can expand by the binomial theorem, integrate, do some algebra, and get a nice answer too. $\endgroup$
    – user217285
    Jul 13, 2015 at 18:44
  • $\begingroup$ Thank you! Will get right on to it then. $\endgroup$
    – MikhaelM
    Jul 13, 2015 at 18:47
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    $\begingroup$ Related:math.stackexchange.com/questions/314602/… $\endgroup$
    – user940
    Jul 13, 2015 at 18:48
  • $\begingroup$ $\displaystyle\lim_{n\to\infty}I_n\sqrt n=\frac{\sqrt\pi}2$ $\endgroup$
    – Lucian
    Jul 13, 2015 at 20:52

5 Answers 5

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Integrate by parts to obtain: $$I_n=\int_0^1 \left(1-x^2\right)^n\,dx=\left(x\left(1-x^2\right)^n\right|_0^1+2n\int_0^1 x^2\left(1-x^2\right)^{n-1}\,dx$$ $$\Rightarrow I_n=2n\int_0^1 \left(x^2-1+1\right)\left(1-x^2\right)^{n-1}\,dx$$ $$\Rightarrow I_n=2nI_{n-1}-2nI_n \Rightarrow I_n=\frac{2n}{2n+1}I_{n-1}$$

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Here is a way you could do it without induction, using the Beta function and the Gamma function. Let $\sqrt{t}=x$, so that $$ I_n=\int_0^1(1-t)^n\frac{t^{1/2}}{2}dt. $$ This now looks like the Beta function, so we have $$ I_n=\frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma ((n+1)+1/2)}. $$ Using the fact that $\Gamma(1/2)=\sqrt{\pi}$, $\Gamma(n+1)=n!$ for positive integer $n$, and $\Gamma((n+1)+1/2)=\frac{(2n+2)!}{4^{n+1}(n+1)!}\sqrt{\pi}$ for positive integer $n$ (all of which can be found in linked wikipedia page), we have $$ I_n=\frac{\sqrt{\pi}n!4^{n+1}(n+1)!}{2(2n+2)!\sqrt{\pi}}=\frac{n!4^{n+1}(n+1)!}{2(2n+2)!}=\frac{n!(n+1)!2^{2n+1}}{(2n+2)!}=\frac{2\cdot4\cdot\dots\cdot 2n}{3\cdot5\cdot\dots 2n+1} $$ Edit: The last equality is somewhat subtle. Note that $(n+1)!\cdot 2^{n+1}$ cancels the even terms in the denominator. The remaining powers of $2$ distribute over the $n!$.

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As Byron Schmuland has pointed out a recursive pattern was developed here. The process in this solution will be a connection to the Beta function.

Consider the integral \begin{align} I_{n} = \int_{0}^{1} \left(1-x^{2}\right)^{n} \, dx. \end{align} Let $t = x^{2}$ to obtain \begin{align} I_{n} &= \frac{1}{2} \, \int_{0}^{1} t^{-\frac{1}{2}} \, (1-t)^{n} \, dt \\ &= \frac{1}{2} \, B\left(\frac{1}{2}, n + 1 \right) \\ &= \frac{\Gamma\left(\frac{3}{2}\right) \, \Gamma\left(n + 1\right)}{\Gamma\left(n + \frac{3}{2}\right)} = \frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}. \end{align}

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For conciseness, let $y:=1-x^2$, and $$\left(xy^n\right)'=y^n-2nx^2y^{n-1}=y^n-2n(1-y)y^{n-1}=(2n+1)y^n-2ny^{n-1}.$$

Then, integrating from $0$ to $1$, $$0=(2n+1)I_n-2nI_{n-1}.$$

Obviously, $I_0=1$.

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  • $\begingroup$ This is essentially the answer by @pranavarora. $\endgroup$
    – user65203
    Jul 13, 2015 at 19:27
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Another simple way would be to substitute $ x = \sin \theta $ and then convert the integral to an integral over the unit circle and finish by using residue theorem...

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