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

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.

  • $\begingroup$ 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


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 $L=\frac{2}{\pi}$.

  • $\begingroup$ 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

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

  • 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

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 $C=-1$.


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

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