Prove that $\lim_{n\rightarrow \infty} \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=0$

Question

Prove that $$\underset{n\rightarrow \infty }{\lim} \ \int_{\epsilon }^{\pi} \frac{\sin(nx)}{nx}dx=0\ ;\ \epsilon>0$$ then use the result to deduce: $$\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=0$$

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My Attempt:

Since $\frac{\sin(nx)}{nx} \leq \frac{1}{n \epsilon} \forall x \in [\epsilon, \pi]$ (of course if we choose $\epsilon$ small enough), it converges uniformly to 0.

Solving first part is trivial, however when it comes to the second one: $$\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\epsilon} \frac{\sin(nx)}{nx}dx+\underset{n\rightarrow \infty }{\lim} \ \int_{\epsilon }^{\pi} \frac{\sin(nx)}{nx}dx$$ I am stuck with the improper integral $\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\epsilon} \frac{\sin(nx)}{nx}dx$. It's obvious that it's equal to 0 but I am facing difficulties in showing that. Help would be appreciated.

Answer 1

Since $$\left\lvert \frac{\sin (nx)}{nx}\right\rvert\leqslant 1,$$

you have

$$\left\lvert \int_0^\epsilon \frac{\sin (nx)}{nx}\,dx\right\rvert \leqslant \epsilon,$$

and therefore

$$\limsup_{n\to\infty} \left\lvert \int_0^\pi \frac{\sin (nx)}{nx}\,dx\right\rvert \leqslant \epsilon.$$

Answer 2

$$\int_0^\pi\frac{\sin(nx)}{nx}dx=\frac1n\int_0^{n\pi}\frac{\sin t}{t}dt\sim_\infty\frac1n \int_0^{\infty}\frac{\sin t}{t}dt$$ and the last integral is convergent ($0$ has a false problem and at $\infty$do an integration by parts to see the convergence).

Answer 3

As $\left\lvert \frac{\sin (nx)}{nx}\right\rvert\leqslant 1$ by Bounded Convergence Theorem we may take the limit inside the integral, so

$\displaystyle\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=\int_{0 }^{\pi} \underset{n\rightarrow \infty }{\lim}\frac{\sin(nx)}{nx}dx=\int_{0}^{\pi} 0 \ dx=0 $

You can use this line of reasoning with your attempt to get $\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\epsilon} \frac{\sin(nx)}{nx}dx+\underset{n\rightarrow \infty }{\lim} \ \int_{\epsilon }^{\pi} \frac{\sin(nx)}{nx}dx= \ \int_{0 }^{\epsilon}\underset{n\rightarrow \infty }{\lim} \frac{\sin(nx)}{nx}dx+ \ \int_{\epsilon }^{\pi} \underset{n\rightarrow \infty }{\lim}\frac{\sin(nx)}{nx}dx=0+0=0$

Answer 4

You have: $$\int_{0}^{\pi}\frac{\sin(nx)}{nx}dx=\frac{1}{n}\int_{0}^{n\pi}\frac{\sin y}{y}=\frac{1}{n}\sum_{j=0}^{n-1}\int_{0}^{\pi}(-1)^j\frac{\sin y}{y+j\pi}dy,$$ so, since $x\,(1-x^2/\pi^2)\geq\sin(x)\geq 0$ when $x\in[0,\pi]$ and $x+(j+1)\pi > x+j\pi\geq 0$, $$0\leq\int_{0}^{\pi}\frac{\sin(nx)}{nx}dx\leq \frac{1}{n}\int_{0}^{\pi}\frac{\sin x}{x}\,dx\leq \frac{1}{n}\int_{0}^{\pi}\left(1-\frac{x^2}{\pi^2}\right)dx=\frac{2\pi}{3n}.$$