How to evaluate $\lim_{n\rightarrow \infty }\left ( \int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}}dx \right ) where\ \forall \ \ n\in \mathbb{N} $? Any Idea how to integrate the Integral in the brackets and than applying limits:
$$\lim_{n\rightarrow \infty }\left ( \int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}} dx\right ) $$
Integral $\int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}}dx $ is the main issue here.
 A: First try to show Uniform convergence of $\frac{\cos(nx)}{x^{2}+n^{2}}$ . We can do this by Weirestrass-M test.
Theorem:- If $\{f_{n}\}$ is a sequence of Riemann integrable functions on $[a,b]$ such that $f_{n}$ converges uniformly to $f$ , then the limiting function $f$ is Riemann integrable on $[a,b]$ and $\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx=\int_{a}^{b}f(x)\,dx$ . That is we can change the order of integral and limit.
We have:-
$$\sup_{x\in[0,2\pi]}|\frac{\cos(nx)}{x^{2}+n^{2}}-0|\leq \sup_{x\in[0,2\pi]} \frac{1}{x^{2}+n^{2}}\leq \frac{1}{n^{2}}$$ .
As $\frac{1}{n^{2}}\to 0$ we have $\frac{\cos(nx)}{x^{2}+n^{2}}$ is uniformly convergent to $0$.
As for each $n\in\mathbb{N}$. $\frac{\cos(nx)}{x^{2}+n^{2}}$ is Riemann integrable in $[0,2\pi]$ as each of them is continuous on this compact interval. So applying the theorem We can interchange the order of limit and integral.
Thus we have
$$
  \lim_{n\to\infty}\int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}}  dx  = \int_{0}^{2\pi}\lim_{n\to\infty}\frac{\cos(nx)}{x^{2}+n^{2}}dx=\int_{0}^{2\pi} 0\,dx = 0$$
A: Since $ 0\le x \le2\pi$, For large $n$ we have $x<<n$ so we can approximate for bounded $x$ values:
$$
  \int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}}  dx\approx   \frac{1}{n^2}\int_{0}^{2\pi} \cos(nx)(1-\frac{x^2}{n^2} ) dx \\\to \frac{1}{n^2}\int_{0}^{2\pi} \cos(nx)dx=\frac1{n^2}\cdot0=0
$$
