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I need to evaluate

$\,\,\,\displaystyle \int \limits_0^{2\pi} \! \sum\limits_{n=1}^\infty \dfrac{\sin nx}{n^3} \, \mathrm{d}x$ and $\displaystyle\int \limits_0^{\pi} \! \sum\limits_{n=1}^\infty \dfrac{\cos nx}{n^2} \, \mathrm{d}x$

I have already proved that the infinite series' are continuous and that the derivative of the first is equal to the second. I'm not sure how to use that information to evaluate the integrals however.

Mathlete
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    Do the series converge uniformly? – Daniel Fischer Nov 10 '13 at 15:52
  • If it converges uniformly do you see further procedure... –  Nov 10 '13 at 15:57
  • The series do converge uniformly do I need to think about Riemann integrability? – Mathlete Nov 10 '13 at 16:04
  • It is not necessary to check for riemann integrability and all... please convince yourself that you can change integral and sum if it is uniformly convergent and then it would be easy to see the result –  Nov 10 '13 at 16:06
  • I'm getting a little confused as to how I would go on from there? – Mathlete Nov 10 '13 at 16:08
  • I said you can surely interchange integral and sum.. Is that a bit clear now? –  Nov 10 '13 at 16:09
  • I understood that part, I just meant the steps after that. Would you then take the limit $n \rightarrow \infty$? – Mathlete Nov 10 '13 at 16:11
  • The series converges uniformly. Then $\int_0^{2\pi} \sum\frac{\sin nx}{n^3}=\sum \int\frac{\sin nx}{n^3}$. – Lei Li Nov 10 '13 at 16:16
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    Yes I understand that, I just don't know how you'd go on after that – Mathlete Nov 10 '13 at 16:17
  • since every $\int\frac{\sin nx}{n^3}=0$, then the sum is also zero. Therefore, the sum of the series in the question is zero. – Lei Li Nov 10 '13 at 18:42

2 Answers2

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Say

$$\int\limits_0^{2\pi}\sum_{n=1}^\infty\frac{\sin nx}{n^3}dx=\sum_{n=1}^\infty\frac1{n^3}\int\limits_0^{2\pi}\sin nx\,dx=\sum_{n=1}^\infty\frac1{n^3}\left(\left.-\frac1n\cos nx\right|_0^{2\pi}\right)=$$

$$=\sum_{n=1}^\infty\frac1{n^3}\cdot 0=0$$

Can you explain the above?

DonAntonio
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Well, something to be said for evaluating one of the sums. I do it here for the cosine: Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$. The result for this series is $x^2/4-\pi x/2+\pi^2/6$ for $x \in [0,2\pi)$. The integral of this series over $[0,\pi]$ is $\pi^3/12 -\pi^3/4 +\pi^3/6 =0$.

For the sine series, integrate the previous sum with respect to $x$, such that the result at $x=0$ is zero; the sum is $x^3/12-\pi x^2/4 +\pi^2 x/6$. You can show that the integral over $[0,2 \pi)$ is zero.

Community
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Ron Gordon
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