5
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
1
$\begingroup$

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}

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

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

and

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

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 $n \to \infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.