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Math Help - Limit (3)

  1. #1
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    Limit (3)

    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).
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  2. #2
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    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.
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  3. #3
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    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.
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