Integral of a function, f, knowing that a trigonometric series converges uniformly to f

Question

I am given the trigonometric series $$\sum_{n=-\infty}^\infty \dfrac{1}{1+n^2}e^{inx}$$ This series converges uniformly to some funciton f(x). f is continous, an even function, 2$\pi$-periodic and its values are real numbers. I need to compute the integrals $$\int_{-\pi}^\pi f(x)dx$$ and $$\int_{-\pi}^\pi f(x)\cos(x)dx$$ which I am not sure how to do. Any help would be much appreciated.

Answer 1

Let $g: \mathbb R \to \mathbb R$ be a bounded function. Then

$$f(x)g(x)= \sum_{n=-\infty}^\infty \dfrac{g(x)}{1+n^2}e^{inx}.$$

The Weierstraß-test shows that the series $\sum_{n=-\infty}^\infty \dfrac{g(x)}{1+n^2}e^{inx}$ converges uniformly on $ \mathbb R$.

Hence

$$\int_{-\pi}^\pi f(x)g(x)dx= \sum_{n=-\infty}^\infty \dfrac{1}{1+n^2} \int_{- \pi}^{\pi}g(x)e^{inx} dx.$$