# Does the series $\sum\limits_{k=1}^{\infty} \int\limits_0^1 \frac{\cos(2 k \pi x) \cos(2 n \pi x)}{k}\,\mathrm dx$ equal $0$ for $n \in \Bbb N$?

Could someone confirm if my answer is correct? $$\sum_{k=1}^{\infty} \int_0^1 \frac{\cos(2 k \pi x) \cos(2 n \pi x)}{k}\,\mathrm dx=0, \quad n\in \mathbb{N}$$

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If $n\neq k$ then the integral is 0, but for $n=k$ you have $\int_0^1 \cos^2(2n\pi x)/n dx \neq 0$. – nullUser May 27 '13 at 16:26
Actually, the sum is equal to $$\frac{\sin 2\pi n}{4\pi n} ( H_{n}+H_{-n} )$$ for any $n \in \Bbb{R}$, where $H_{s}$ is the harmonic number. – Sangchul Lee May 27 '13 at 18:58

$$I:=\int\limits_0^1\frac{\cos^22n\pi x}{n}dx$$
$$u:=2n\pi x\implies dx=\frac{du}{2n\pi}\implies$$
$$I=\frac1{2n^2\pi}\int\limits_0^{2n\pi}\cos^2u\,du=\left.\frac1{2n^2\pi}\left(\frac{u+\sin u\cos u}2\right)\right|_0^{2n\pi}=\frac1{2n^2\pi}\left(\frac{2n\pi}2\right)=\frac1{2n}$$
I've made the $I$s in your answer italic, because they looked a lot like $1$s. I hope you don't mind too much. – Lord_Farin May 27 '13 at 16:40
Thanks for the editing but I'd rather have them as they were. I'll leave them as you did because I'd like to read some other members' opinions on this, but I think that from the two options $\,\text{I}\;,\;I\,$ , the former one is better. – DonAntonio May 27 '13 at 16:44