Limit (3)

NonCommAlg · #1 $Old$ July 5th, 2009, 02:33 AM

For any positive integers $m,n$ define $f(m,n)=\int_{\frac{1}{m}}^{\frac{\pi}{2}} \frac{\sin(nx)}{\sin x} \ dx.$ Evaluate $L_1=\lim_{m\to\infty} \lim_{n\to\infty} f(m,n)$ and $L_2=\lim_{n\to\infty} \lim_{m\to\infty} f(m,n).$

halbard · #2 $Old$ July 5th, 2009, 03:17 PM

The function $\frac1{\sin x}$ is in $\textrm C[{\textstyle\frac1m},{\textstyle\frac\pi2}]\subset\mathrm L^1[{\textstyle\frac1m},{\textstyle\frac\pi2}]$ , and by the Riemann-Lebesgue lemma $\lim_{n\to\infty}\int_{1/m}^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx=0$ . Therefore $\lim_{m\to\infty}\lim_{n\to\infty}f(m,n)=0$ .

The function $\frac{\sin nx}{\sin x}$ is in $\mathrm C[0,{\textstyle\frac\pi2}]$ (with $f(0)=n$ ) and is improperly Riemann integrable there, so $\lim_{m\to\infty}\int_{1/m}^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx=\int_0^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx$ .

The function $\frac1x-\frac1{\sin x}$ (with $f(0)=0$ ) is in $\mathrm C[0,{\textstyle\frac\pi2}]$ , and is improperly Riemann integrable and Lesbesgue integrable there.

By the Riemann-Lebesgue lemma, $\lim_{n\to\infty}\int_0^{\pi/2}\left(\frac{\sin nx}{x}-\frac{\sin nx}{\sin x}\right)\mathrm dx=0$ .

Thus $\lim_{n\to\infty}\int_0^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx=\lim_{n\to\infty}\int_0^{\pi/2}\frac{\sin nx}x\mathrm dx=\lim_{n\to\infty}\int_0^{n\pi/2}\frac{\sin x}x\mathrm dx=\int_0^\infty\frac{\sin x}x\mathrm dx=\frac\pi2$ . Therefore $\lim_{n\to\infty}\lim_{m\to\infty}f(m,n)=\frac\pi2$ .

NonCommAlg · #3 $Old$ July 5th, 2009, 05:56 PM

Quote:

Originally Posted by halbard $View Post$

The function $\frac1{\sin x}$ is in $\textrm C[{\textstyle\frac1m},{\textstyle\frac\pi2}]\subset\mathrm L^1[{\textstyle\frac1m},{\textstyle\frac\pi2}]$ , and by the Riemann-Lebesgue lemma $\lim_{n\to\infty}\int_{1/m}^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx=0$ . Therefore $\lim_{m\to\infty}\lim_{n\to\infty}f(m,n)=0$ .

The function $\frac{\sin nx}{\sin x}$ is in $\mathrm C[0,{\textstyle\frac\pi2}]$ (with $f(0)=n$ ) and is improperly Riemann integrable there, so $\lim_{m\to\infty}\int_{1/m}^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx=\int_0^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx$ .

The function $\frac1x-\frac1{\sin x}$ (with $f(0)=0$ ) is in $\mathrm C[0,{\textstyle\frac\pi2}]$ , and is improperly Riemann integrable and Lesbesgue integrable there.

By the Riemann-Lebesgue lemma, $\lim_{n\to\infty}\int_0^{\pi/2}\left(\frac{\sin nx}{x}-\frac{\sin nx}{\sin x}\right)\mathrm dx=0$ .

Thus $\lim_{n\to\infty}\int_0^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx=\lim_{n\to\infty}\int_0^{\pi/2}\frac{\sin nx}x\mathrm dx=\lim_{n\to\infty}\int_0^{n\pi/2}\frac{\sin x}x\mathrm dx=\int_0^\infty\frac{\sin x}x\mathrm dx=\frac\pi2$ . Therefore $\lim_{n\to\infty}\lim_{m\to\infty}f(m,n)=\frac\pi2$ .

beautiful! my solution is the same as yours except that i proved $\lim_{n\to\infty}\int_{1/m}^{\pi/2}\frac{\sin nx}{\sin x}\mathrm dx=0$ and $\lim_{n\to\infty}\int_0^{\pi/2}\left(\frac{\sin nx}{x}-\frac{\sin nx}{\sin x}\right)\mathrm dx=0$ without using the Riemann-Lebesgue lemma.

so i'll leave that open to whoever wants to try it.