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If $a>0,$ show that $\lim \limits_{n \to \infty} \int_{ a}^{\pi} \frac{\sin nx}{nx} \mathrm dx =0,$ what happens if $a=0.$
The first part is easy since the integrand uniformly converges to zero on $[a, \pi]$ and for each $n, \; f_n(x)= \frac {\sin nx}{nx}$ is continuous on $[a,\pi].$
But for $a=0$ I don't understand how to proceed. I was thinking of using bounded convergence theorem but that required uniform boundedness of the integrand . Please give me a hint. Thanks in advance.

Normal
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2 Answers2

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We have that $$\int_{0}^{\pi }\frac{\sin nx}{nx} \mathrm dx=\int_{0}^{n\pi }\frac{\sin nx}{nx} \mathrm {1\over n}dnx={1\over n }\int_{0}^{n\pi} {\sin x\over x}dx$$also we know that $$\int_0^{\infty}{\sin x\over x}dx={\pi \over 2}$$according to Integration of Sinc function therefore $$\lim _{n\to \infty}\int_{0}^{\pi }\frac{\sin nx}{nx} \mathrm dx=0$$

Mostafa Ayaz
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  • Thank you .So $ \lim_{n\to \infty} \int_{0}^{\pi} \frac{\sin nx}{nx} $=$( lim_{n\to \infty} {1 \over n})(lim_{n\to \infty} \int_{0}^{ n\pi}{\sin x over x}) dx $ =0×(pi/2)=0. Is it what you did? – Normal Nov 23 '18 at 20:05
  • You're welcome. Also could be explained that way. Yes... – Mostafa Ayaz Nov 23 '18 at 20:10
  • Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here. – Normal Nov 23 '18 at 21:18
  • No mind at all :) Notice that i concluded that $$0\le \lim \int_{0}^{\pi }\frac{\sin nx}{nx} \mathrm dx= \lim {1\over n }\int_{0}^{n\pi} {\sin x\over x}dx\le\lim {1\over n }\int_{0}^{\infty} {\sin x\over x}dx=\lim {\pi \over 2n}=0$$ and used squeeze theorem – Mostafa Ayaz Nov 23 '18 at 21:21
  • Thanks for answering though I didn't understand the line where you wrote lim $ {1 \over n} \int_{0}^{n\pi} { \sin x \over x} dx \le lim {1 \over n} \int_{0}^{\infty} {\sin x \over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero . – Normal Nov 23 '18 at 22:31
  • Use this theorem that $\lim f(x)g(x)=\lim f(x)\lim g(x)$ since both limits exist and are bounded: one is zero and the other is finite and positive – Mostafa Ayaz Nov 24 '18 at 07:02
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For all $x$, $|f_n(x)|\le1$, hence

$$\left|\int_0^{a}f_n(x)dx\right|\le (a-0) \sup_{[0,a]}|f_n|\le a$$

Therefore

$$\left|\int_0^\pi f_n(x)dx\right|=\left|\int_0^a f_n(x)dx+\int_a^\pi f_n(x)dx\right|\le a+\left|\int_a^\pi f_n(x)dx\right|$$

Now, for all $\epsilon>0$, let $a=\epsilon$, and the remaining integral term tends to $0$ as $n\to0$ (as $a$ is fixed as soon as $\epsilon$ is fixed), so there is a $N$ such that for all $n>N$, $\left|\int_a^\pi f_n(x)dx\right|<\epsilon$.

Hence, for all $\epsilon>0$, there is a $N$ such that for all $n>N$, $\left|\int_0^\pi f_n(x)dx\right|<2\epsilon$.

Hence $\int_0^\pi f_n(x)dx\to0$ as $n\to\infty$.

Jean-Claude Arbaut
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  • Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| \le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ? – Normal Nov 23 '18 at 22:19
  • @Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{\infty}$. See also https://en.wikipedia.org/wiki/Sinc_function – Jean-Claude Arbaut Nov 23 '18 at 22:21
  • Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $ – Normal Nov 23 '18 at 22:38
  • @Normal Yes, it also works. But you're not really a beginner then :) – Jean-Claude Arbaut Nov 23 '18 at 22:41
  • One more question what if I choose an $ \epsilon $ different from a . – Normal Nov 23 '18 at 23:05
  • @Normal It's the other way around: the goal is to get a statement "$\forall \epsilon>0,\exists N,\forall n>N$... your integral is less that $\epsilon$" (or a constant multiple of $\epsilon$), in order to prove it tends to $0$. So first you assume there is some $\epsilon>0$. then you decide to cut the integral. You could take $a=g(\epsilon)$ for some (not completely arbitrary) function of $g$, but the simplest is $a=\epsilon$. – Jean-Claude Arbaut Nov 23 '18 at 23:11
  • No sorry but I didn't get it ...thanks anyway :) – Normal Nov 23 '18 at 23:27