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

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.
 A: 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
$$\limsup_{n\to\infty} \left\lvert \int_0^\pi \frac{\sin (nx)}{nx}\,dx\right\rvert \leqslant \epsilon.$$
A: $$\int_0^\pi\frac{\sin(nx)}{nx}dx=\frac1n\int_0^{n\pi}\frac{\sin t}{t}dt\sim_\infty\frac1n \int_0^{\infty}\frac{\sin t}{t}dt$$
and the last integral is convergent ($0$ has a false problem and at $\infty$do an integration by parts to see the convergence).
A: 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 $\underset{n\rightarrow \infty }{\lim} \  \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=\underset{n\rightarrow \infty }{\lim} \  \int_{0 }^{\epsilon} \frac{\sin(nx)}{nx}dx+\underset{n\rightarrow \infty }{\lim} \  \int_{\epsilon }^{\pi} \frac{\sin(nx)}{nx}dx= \  \int_{0 }^{\epsilon}\underset{n\rightarrow \infty }{\lim} \frac{\sin(nx)}{nx}dx+ \  \int_{\epsilon }^{\pi} \underset{n\rightarrow \infty }{\lim}\frac{\sin(nx)}{nx}dx=0+0=0$
A: 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 $x+(j+1)\pi > x+j\pi\geq 0$,
$$0\leq\int_{0}^{\pi}\frac{\sin(nx)}{nx}dx\leq \frac{1}{n}\int_{0}^{\pi}\frac{\sin x}{x}\,dx\leq \frac{1}{n}\int_{0}^{\pi}\left(1-\frac{x^2}{\pi^2}\right)dx=\frac{2\pi}{3n}.$$
