Is it posible to integrate $\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n}\sin(nx)$ term to term when $x \in [-\pi,\pi]$?

Question

Let $f(x)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n}\sin(nx)$. I want to calculate its integral. Can I integral it term to term?

$\displaystyle\int\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n}\sin(nx)dx=\displaystyle\sum_{n=1}^{\infty}\int[(-1)^{n+1}\frac{2}{n}\sin(nx)]dx$

$x\in[-\pi,\pi]$

Answer 1

You can use Dirichlet's theorem to prove the series uniformly convergence. Futhermore, each term integrable. Thus, you can integral it term to term.