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I saw this integral in a paper on hypergeometric functions: $$S(n)=\int_{0}^{\pi/2}\sin(t)^{2n+3}dt=\frac{4^n(2n+2)}{(2n+3)(2n+1){2n\choose n}}\;\;\;\;\;\;\;\;\;\;\;(1)$$ I tried to prove it and got this far.

Given $$\int_{0}^{\pi/2}\sin^{\alpha}tdt=\frac1{2}\beta(\frac{\alpha+1}{2},\frac1{2})$$ We arrive at $$S(n)=\frac1{2}\beta(n+2,\frac1{2})$$ $$S(n)=\frac{\sqrt{\pi}\,\Gamma(n+2)}{2\Gamma(n+5/2)}$$ $$S(n)=\frac{\sqrt{\pi}}{2}\frac{n(n+1)}{(n+3/2)(n+1/2)}\frac{\Gamma(n)}{\Gamma(n+1/2)}$$ $$S(n)=\frac{2n+2}{(2n+3)(2n+1)}n\beta(n,\frac{1}{2})$$ And assuming that $(1)$ is true, we get $$\beta(n,1/2)=\beta(1/2,n)=\frac{4^n}{n{2n\choose n}}\;\;\;\;\;\;\;(2)$$ But I do not know how to prove $(2)$.

I tried to prove it by starting with $$\beta(1/2,n)=\frac{\Gamma(1/2)\Gamma(n)}{\Gamma(n+1/2)}$$ But I do not know how to prove $$\frac{\Gamma(1/2)\Gamma(n)}{\Gamma(n+1/2)}=\frac{4^n}{n{2n\choose n}}$$

clathratus
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  • Just set $t=\arcsin\sqrt{u}$ to convert $S(n)$ into a value of the Beta function. It is worth to recall that $$\beta(1/2,n)= \frac{4^n}{n\binom{2n}{n}}.$$ – Jack D'Aurizio Oct 24 '18 at 16:36
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    I guess the paper is the one born from https://math.stackexchange.com/questions/2123298/what-is-4f-3-left1-1-1-frac32-frac52-frac52-frac521-r/2130679#2130679 – Jack D'Aurizio Oct 24 '18 at 16:47
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    @JackD'Aurizio I know how to get from the original integral to $\frac{2n+2}{(2n+3)(2n+1)}n\beta(1/2,n)$, but my point of confusion is on the derivation of $$\beta(1/2,n)=\frac{4^n}{n{2n\choose n}}$$ – clathratus Oct 24 '18 at 16:50
  • The integrals $\int_{0}^{\pi/2}\sin(\theta)^n,d\theta$ can be simply computed by integration by parts or the lemma $\int_{0}^{2\pi} e^{mi\theta},d\theta = 2\pi\delta(m)$. They are typically involved in the derivation of Wallis' product. – Jack D'Aurizio Oct 24 '18 at 16:52
  • @JackD'Aurizio and that's the Dirac delta? – clathratus Oct 24 '18 at 16:54
  • @JackD'Aurizio I edited the question a lot. Hopefully it's easier to understand now. – clathratus Oct 24 '18 at 17:32
  • This follows easily from reduction formula for $I_n=\int_{0}^{\pi/2}\sin ^{n} x, dx$ given by $nI_n=(n-1)I_{n-2}$. – Paramanand Singh Nov 01 '18 at 23:11
  • @ParamanandSingh for all $n$? or just $n\in\Bbb N$ – clathratus Nov 01 '18 at 23:12
  • @clathratus: it is generally used for $n\in\mathbb{N}$, but the proof via integration by parts is valid for any $n$ with $n \geq 1$. For the current question I think $n\in\mathbb{N}$. – Paramanand Singh Nov 02 '18 at 02:34

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A nice method to integrate $\sin t$ to an odd power is to keep one factor of sine, and convert all the rest to cosine (using $\sin^2 t = 1-\cos^2 t$), then substitute $x = \cos t$, $dx = -\sin t\;dt$. $$ \int_0^{\pi/2} (\sin t)^{2n+3} dt = \int_0^{\pi/2} (1-\cos^2 t)^{n+1} \sin t\;dt = \int_0^1 (1-x^2)^{n+1} \;dx $$ the integral of a polynomial, which you already know how to do.

This way, there is no need to know about gamma or beta functions.

Integrating the polynomial, we get $$ \sum_{j=0}^{n+1}\frac{(-1)^j\binom{n+1}{j}}{2j+1} $$ So we have to do some combinatorics to reach your answer $$ \frac{(2n+2)!!}{(2n+3)!!} = \frac{4^n (n!)^2(2n+2)}{(2n+1)(2n+3)(2n)!} $$

GEdgar
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