Proving $\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx=0 $ $$\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx=0 $$


I consider if it can be solved by other methods without Riemann lemma.
I try my best to do it as follow:


  Let $t=nx$
  $$\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx$$
  \begin{eqnarray} 
&amp;=&amp;\lim_{n\to \infty}\int_0^{n\pi}\frac{\sin\left(t\right)}{1+(\frac{t}{n})^2}d\frac{t}{n}\\
&amp;=&amp;\lim_{n\to \infty}\int_0^{n\pi}\frac{n\sin\left(t\right)}{n^2+t^2}dt\\
&amp;=&amp;\lim_{n\to \infty}\sum_{k=0}^{k=n-1}\int_{k\pi}^{(k+1)\pi}\frac{n\sin\left(t\right)}{n^2+t^2}dt\\
\end{eqnarray} 


Then I can not go on. Who can tell me how to prove it? Thank you.
Integration by parts is enough:

$$\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx =\left.\frac{\frac{1}{n}(1-\cos(nx))}{1+x^2}\right|_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\frac{2x\,(1-\cos(nx))}{(1+x^2)^2}\,dx,\tag{1}$$
so:
$$\left|\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx\right|\leq\frac{1}{n}\left(\frac{2}{\pi^2+1}+\frac{2\pi^2}{\pi^2+1}\right)=\frac{2}{n}.\tag{2}$$
Using the Cauchy-Schwarz inequality to bound the second integral in $(1)$, it is possible to improve the last inequality up to:
$$\left|\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx\right|\leq\frac{1}{n}\left(\frac{2}{\pi^2+1}+\sqrt{\frac{3\pi}{2}\int_{0}^{\pi}\frac{4x^2}{(1+x^2)^4}\,dx}\right)&lt;\frac{14}{9n}.\tag{3}$$
For each integral of the sum $|I_k| = \left|\displaystyle \int_{k\pi}^{(k+1)\pi} \dfrac{n\sin t}{n^2+t^2}dt\right| \leq \displaystyle \int_{k\pi}^{(k+1)\pi} \dfrac{1}{n}dt = \dfrac{\pi}{n}$. Taking the limit $n \to \infty$ we have: $I_k \to 0$ , $\forall k$. You're done.