I want to prove that $$\lim_{n\to \infty }\int_{\mathbb R}\frac{\sin(nx)}{nx}dx=\int_{\mathbb R}\lim_{n\to \infty }\frac{\sin(nx)}{nx}dx,$$ i.e. to majorate $|\frac{\sin(nx)}{nx}|$ by an integrable function over $\mathbb R$ that doesn't depend on $n$. Any idea ?
Prove $\lim\limits_{n\to \infty }\int_{\mathbb R}\frac{\sin(nx)}{nx}dx=\int_{\mathbb R}\lim\limits_{n\to \infty }\frac{\sin(nx)}{nx}dx$
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real-analysis
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2You can't majorate $|\frac{\sin(nx)}{nx}|$ by an integrable function: it's not integrable over $\mathbb{R}$. – 2017-01-01
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0@OlivierOloa: So, how can I do ? – 2017-01-01
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0Are you sure the equality holds? – 2017-01-01
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0@OlivierOloa: Yes, since $$\lim_{n\to \infty }\int_{\mathbb R}\frac{\sin(nx)}{nx}d \mu(x)=\mu(\{0\})=\int_{\mathbb R}\boldsymbol 1_{x=0}(x)d\mu(x).$$ – 2017-01-01
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0Ok, it's what you want to prove. – 2017-01-01
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1The first integrand in the title is not Lebesgue $\mathbb{R}$-integrable. – 2017-01-01
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2You can prove that both sides are equal, but this will not (at least directly) follow from dominated convergence theorem. It simply follows from the fact that $$ \int_{-\infty}^{\infty} \operatorname{sinc}(nx) \, dx = \frac{\pi}{n} \qquad \text{and} \qquad \lim_{n\to\infty} \operatorname{sinc}(nx) = \mathbf{1}_{\{x = 0\}}, $$ where $\operatorname{sinc}(x) = \sin x / x$ with singularity at $x = 0$ removed by continuation. So both sides vanish and they are equal. – 2017-01-01
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0The integral exists as an improper Riemann integral. Enforcing the substitution $y=nx$ clearly shows the limit of the integral is $0$. And obviously the point-wise limit of the integrand is $0$ almost everywhere. So, the equality holds. – 2017-01-01
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0@MSE observe that @ SangchulLee has given the route. – 2017-01-01
1 Answers
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This is my attempt if not correct please pointout the mistakes. I could not comment as it overflowed with characters.
You say that RHS is $0$ since $sin(nx)$ is bounded and $\frac{1}{nx} \rightarrow 0$. Then say that $$\int_{-\infty}^{\infty} \frac{sin(nx)}{nx}dx =2\int_{0}^{\infty} \frac{sin(nx)}{nx}dx = (2/n)\int_{0}^{\infty} \frac{sin(t)}{t}dt =(2/n)\int_{0}^{1} \frac{sin(t)}{t}dt + (2/n)\int_{1}^{\infty} \frac{sin(t)}{t}dt$$
$ \int_{0}^{1} \frac{sin(t)}{t}dt $ is bounded since $\frac{sin(t)}{t} $ is continuous in $[0,1]$ and
$\int_{1}^{\infty} \frac{sin(t)}{t}dt$ will converge using abel dirichlets theorem/test. Hence then given $ 1 >\epsilon >0$ you can choose $N$ accordingly so that you get $\forall n\geq N $ we have $$|\int_{-\infty}^{\infty} \frac{sin(nx)}{nx}dx| < \epsilon$$