How do you solve this limit involving definite integration?

Question

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

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.

Answer 1

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

Answer 2

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

Answer 3

This is to supply Jack D'Aurizio's claim that $$ \lim_{r\to +\infty}\frac{\int_{0}^{\pi/2}x^{r+1}\sin(x)\,dx}{\int_{0}^{\pi/2}x^r\sin(x)\,dx} = \frac{\pi}{2} $$

There are interesting solutions to this claim. One that I like may be obtained using the result in this posting:

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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 $$\frac{\alpha_{p+1}}{\alpha_p}\xrightarrow{p\rightarrow\infty}\|f\|_\infty$$

Notes:

• The proof is based on Holder's inequality along with the well known limit $\lim_{p\rightarrow\infty}\|f\|_p=\|f\|_\infty$.
• It is possible to avoid resorting to Lebesgue integration by just using Holder inequalities for Riemann integrals. After all, the problem at hand involves nice continuous functions in a closed bounded interval.

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For the question at hand, consider the measure space $([0,\pi/2],\mathscr{B}([0,\pi/2]),\mu)$, where $\mu(dx)=\mathbb{1}_{[0,\pi/2]}(x)\,\sin(x)\,dx$. The function $f(x)=x$ is bounded in this space and $\|f\|_\infty=\frac{\pi}{2}$.