I wish to show $$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$$

I've tried substitution and integration by parts to get a recursive formula for the integral, but so far it hasn't worked. Any help would be appreciated, thanks.

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    $\begingroup$ Use $\sin^n x = (\sin^{n-2}x)(\sin^2 x) = (\sin^{n-2}x)(1-\cos^2 x)$. Then use parts to end up with $\sin^{n-2}x \cos^{n-2}x$ which you collapse to a single trig function via the double angle formula for sine. See the 16th line here ( en.wikipedia.org/wiki/… ) for help on the parts step. $\endgroup$ – Eric Towers Jun 19 '14 at 7:11

I will present you the main steps to a final solution and the details on each step can be your job.

First notice that a simple substiution $n=2k$ gives

$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \lim_{k\rightarrow \infty} \int_0^{\pi/2} 2^{2k} \sqrt{2k} \sin^{2k}(x) \cos^{2k-2}(x) \; dx$$ Now elaborating with the new expression in the following way we have

$$2^{2k-1} \sqrt{2k}\cdot2\int_0^{\pi/2}\sin^{2(k+1/2)-1}(x) \cos^{2(k-1/2)-1}(x) \; dx$$

\begin{eqnarray} &=& 2^{2k-1}\sqrt{2k}\cdot\beta(k+1/2,k-1/2) \\ &=& 2^{2k-1}\sqrt{2k}\frac{\Gamma(k+1/2)\Gamma(k-1/2)}{\Gamma(2k)} \\ &=& 2^{2k}\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{\Gamma(k+1/2)^2}{\Gamma(2k+1)} \\ &=& 2^{2k}\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{(\frac{(2k-1)!!\sqrt{\pi}}{2^{k}})^{2}}{(2k)!} \\ &=& \pi\cdot\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{(2k-1)!!}{(2k)!!} \\ &=& \pi\sqrt{\frac{2k}{2k+1}}\frac{2k}{2k-1}\cdot\sqrt{\frac{(2k-1)!!^2(2k+1)}{(2k)!!^2}} \\ &=& \;\pi\sqrt{\frac{2k}{2k+1}}\frac{2k}{2k-1}\sqrt{\prod_{j=1}^{k}\frac{(2j-1)}{(2j)}\frac{(2j+1)}{(2j)}} \\ & \rightarrow & \pi\cdot\sqrt{1}\cdot1\sqrt{\prod_{j=1}^{\infty}\frac{(2i-1)}{(2i)}\frac{(2i+1)}{(2i)}} \\ &=&\pi\cdot\sqrt{\frac{2}{\pi}} \\ &=& \sqrt{2\pi},\;k\rightarrow \infty \end{eqnarray}

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    $\begingroup$ Does this work if $n$ is odd? $\endgroup$ – Paramanand Singh Jun 20 '14 at 3:04
  • $\begingroup$ Indeed it does. We actually get nice things to work with since the expression simplifies to $$2^{2n}\sqrt{2n+1}\frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)}$$ $\endgroup$ – TheOscillator Jun 20 '14 at 10:10
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    $\begingroup$ Thats cool! by the way in your above comment you should replace $n$ by $k$ so that you handle the $n = 2k + 1$ case. case $n = 2k$ is already handled in your answer. I wish i could upvote one more time! $\endgroup$ – Paramanand Singh Jun 20 '14 at 10:22
  • $\begingroup$ It took a little bit of hacking but I got the equations to align nicely. Please in the future DO NOT write your answers the way you did initially. It was like Where's Waldo met Jackson Pollock. $\endgroup$ – Cameron Williams Jun 21 '14 at 2:59

If we write

$$ \int_0^{\pi/2} \sin^n x \cos^{n-2} x\,dx = \int_0^{\pi/2} \cos^{-2} x \exp\Bigl[n\log(\sin x \cos x)\Bigr]\,dx $$

and note that the quantity $\log(\sin x \cos x)$ achieves a maximum over the interval of integration at $x = \pi/4$, then by calculating

$$ \log(\sin x \cos x) = -\log 2 - 2 \left(x-\frac{\pi}{4}\right)^2 + O\left(\left(x-\frac{\pi}{4}\right)^4\right) $$


$$ \cos^{-2} x = 2 + O\left(x-\frac{\pi}{4}\right) $$

as $x \to \pi/4$ we may appeal to the Laplace method to conclude that

$$ \int_0^{\pi/2} \sin^n x \cos^{n-2} x\,dx \sim \int_{-\infty}^{\infty} 2 e^{-n\log 2-2ny^2}\,dy = \frac{\sqrt{2\pi}}{2^n\sqrt{n}} $$

as $n \to \infty$.


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