## lack of uniform convergence

How can I justify that $\sum_{n=1}^{\infty}\frac{1}{n}\int_{2\pi}^{\infty} \frac{\sin nx}{x}\ dx =\int^{\infty}_{2\pi}\frac{1}{x}\sum_{n=1}^{\infty }\frac{\sin nx}{n}\ dx$ ?

What I want to do is the following:

$\sum_{n=1}^{\infty}\frac{1}{n}\int_{2\pi}^{\infty} \frac{\sin nx}{x}\ dx =\int^{\infty}_{2\pi}\frac{1}{x}\sum_{n=1}^{\infty }\frac{\sin nx}{n}\ dx$

$= \sum_{k=1}^{\infty} \int_{2k \pi}^{2(k+1) \pi} \frac{1}{x} \Big(\frac{\pi-(x-2k \pi)}{2} \Big) \ dx$

since $f(x) = \frac{\pi -x}{2} = \sum_{n=1}^{\infty} \frac{\sin nx}{n} \ \text{for} \ (0,2 \pi]$ and $f(x+2 \pi)$ otherwise