# Find $\int_0^{2\pi}g(x)\space dx$ where $g(x)$ is the sum of the series

$$\sum^\infty_{n=1}\frac{\cos(2nx)}{n^3}$$ Let $$g(x)$$ denote the sum of the series.

Find $$\int_0^{\frac{\pi}{2}}g(x)\space dx.$$

I have already shown that the series converges uniformly on $$\mathbb{R}$$.

I used this theorem:

$$\int^b_af=\int^b_a\sum^\infty_{n=1}f_n=\sum^\infty_{n=1}\int^b_af_n$$

My method was too use the theorem to swap the integral and sum around, in order to find the pointwise on $$x\in [0, 2\pi]$$. Then using the same theorem find the pointwise, and integrate it.

I found the pointwise limit $$f$$ to be $$0$$, as $$\lim\limits_{n \to \infty} \frac{1}{n^3}\to 0$$.

Then using the pointwise limit to find the integral:$$\int^{\frac{\pi}{2}}_00\space dx=0$$

I think I've gone wrong here somewhere, either by using the wrong method or the wrong application of the correct method. I would be grateful if someone could point some potential mistakes for me.

• The limit of the summands is zero but that's not the limit of the sum itself. – Ian Jan 11 at 17:22
• $$\sum_{n\geq 1}\frac{\cos(2nx)}{n^3}$$ is not constantly zero, but its integral over $[0,2\pi]$ clearly equals zero by symmetry. – Jack D'Aurizio Jan 11 at 17:22

You have done well! Now $$\int_0^{2\pi}\frac{\cos(2nx)}{n^3}= \Bigl[\frac{\sin(2nx)}{2n^4}\Bigr]_0^{2\pi}=0$$ Now use that $$\int_0^{2\pi}\sum_{n>0}f_n=\sum_{n>0}\int_0^{2\pi}f_n$$ Nothing else is needed.
• @james15c It's $\int_0^{2\pi}f_n$ – egreg Jan 11 at 17:52