Finding $\int_{-\pi}^{\pi} f(x)dx$, where $f(x) = \sum_{k=1}^{\infty} \frac{\cos{(k^2x)}}{k^p}$, with $p > 1$ and $x \in \mathbb{R}$

Question

I'm not sure how to evaluate this integral. I've shown that $\sum_{k=1}^{\infty} \frac{\cos{(k^2x)}}{k^p}$ converges uniformly on $\mathbb{R}$ via the Weierstrass M-test, but I don't know how to proceed. I can't find what it converges to, so I don't really know where to go from here. What am I missing?

Answer 1

Let $f_n(x) = \sum_{k=1}^n \frac{\cos(k^2x)}{k^p}$. Then $|f_n(x)|\leqslant g(x) := \sum_{k=1}^\infty \frac1{k^p} =\zeta(p)<\infty$, so by the dominated convergence theorem, \begin{align} \int_{-\pi}^\pi \lim_{n\to\infty} f_n(x)\ \mathsf dx &= \lim_{n\to\infty}\int_{-\pi}^\pi f_n(x)\ \mathsf dx\\ &=\lim_{n\to\infty}\int_{-\pi}^\pi \sum_{k=1}^n \frac{\cos(k^2x)}{k^p}\ \mathsf dx\\ &=\sum_{n=1}^\infty \int_{-\pi}^\pi \frac{\cos(n^2x)}{n^p}\ \mathsf dx\\ &=\sum_{n=1}^\infty \frac1{n^p} \int_{-\pi}^\pi \cos(n^2 x)\ \mathsf dx\\ &=0, \end{align} since $\cos$ is an even function.