How can I justify that $\displaystyle \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:


$\displaystyle \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 $


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


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