7
$\begingroup$

$$ \lim \limits_{r \to \infty} \frac {r^C \int_0^{\frac{\pi}{2}} x^r \sin(x)\, dx}{\int_0^{\frac{\pi}{2}} x^r \cos(x)\, dx} = L$$

Find the value of $\pi L - C$, given that $C\in\mathbb{R}$ and $L>0$.

My approach:

I tried to apply integration by parts to both the numerator and denominator to get a recurring relation, hoping to cancel something off, but to no avail. I'm not getting any other method to solve it, so any help will be appreciated.

$\endgroup$
11
  • $\begingroup$ Won't the limit depend on the value of $C$? $\endgroup$
    – najayaz
    Oct 25, 2015 at 15:20
  • $\begingroup$ That's the thing. You're supposed to get the value of C so that the limit is a finite quantity (which is equal to L, which too you have to find). $\endgroup$ Oct 25, 2015 at 15:21
  • $\begingroup$ That's just too much work for a single question. $\endgroup$
    – najayaz
    Oct 25, 2015 at 15:23
  • $\begingroup$ The integrals come out in terms of hypergeometric functions so I wouldn't spend much time on that. The answer is 3 but I have no idea how to do it without cheating. $\endgroup$
    – Ian Miller
    Oct 25, 2015 at 15:23
  • $\begingroup$ @G-Man I know, but I think it's a really well thought of question. $\endgroup$ Oct 25, 2015 at 15:23

3 Answers 3

12
$\begingroup$

I probably have a simple solution that many missed, through a probabilistic/distributional approach. It is quite trivial that:

$$ \lim_{r\to +\infty}\frac{\int_{0}^{\pi/2}x^{r+1}\sin(x)\,dx}{\int_{0}^{\pi/2}x^r\sin(x)\,dx} = \frac{\pi}{2} $$ since the integrand functions in the numerator/denominator get more and more concentrated around the right endpoint as $r$ increases, and their ratio at $x=\frac{\pi}{2}$ is exactly $\frac{\pi}{2}$. By using integration by parts, we have: $$ \lim_{r\to +\infty}\frac{(r+1)\int_{0}^{\pi/2}x^{r}\cos(x)\,dx}{\int_{0}^{\pi/2}x^r\sin(x)\,dx} = \frac{\pi}{2}$$ hence the given limit is finite iff $C=-1$ and in such a case $L=\frac{2}{\pi}$.

$\endgroup$
2
  • $\begingroup$ again to slow for @Jack (+1) $\endgroup$
    – tired
    Oct 25, 2015 at 16:04
  • $\begingroup$ This shortcut is pretty neat. Thanks :D $\endgroup$ Oct 26, 2015 at 20:00
1
$\begingroup$

This is 2011 putnam A3 problem,you can see some solution :http://www.artofproblemsolving.com/community/c7h449984p2531777

$\endgroup$
1
  • 3
    $\begingroup$ I'm not familiar with the norm of this community, but if this isn't a link-only answer, then what is it? $\endgroup$
    – M.A.R.
    Oct 25, 2015 at 20:57
0
$\begingroup$

One may solve the problem go the OP by an application of the following result

Proposition: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space and $f\in L_\infty$. Define $\alpha_p =\int_X |f|^p\,d\mu$. Then $$\frac{\alpha_{p+1}}{\alpha_p}\xrightarrow{p\rightarrow\infty}\|f\|_\infty$$

to the case $([0,\pi/2],\mathscr{B}([0,\pi/2]),\mu)$, where $\mu(dx)=\mathbb{1}_{[0,\pi/2]}(x)\,\sin(x)\,dx$, and $f(x)=x$. In this case, after an application of integration by parts, we obtain $$ \lim_{r\to \infty}\frac{\int_{0}^{\pi/2}x^{r+1}\sin x\,dx}{\int_{0}^{\pi/2}x^r\sin x\,dx} = \lim_{r\to\infty} \frac{(r+1)\int^{\pi/2}_0 x^r\cos x\,dx}{\int^{\pi/2}_0 x^r\sin x\,dx}=\|f\|_\infty=\frac{\pi}{2} $$

This gives possible values $L=\frac{2}{\pi}$ and $C=-1$.


Note:

  • A proof of the Proposition above can be found here. The proof is based on Holder's inequality along with the well known limit $\lim_{p\rightarrow\infty}\|f\|_p=\|f\|_\infty$.
$\endgroup$

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.