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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.

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    $\begingroup$ Do the series converge uniformly? $\endgroup$ – Daniel Fischer Nov 10 '13 at 15:52
  • $\begingroup$ If it converges uniformly do you see further procedure... $\endgroup$ – user87543 Nov 10 '13 at 15:57
  • $\begingroup$ The series do converge uniformly do I need to think about Riemann integrability? $\endgroup$ – Mathlete Nov 10 '13 at 16:04
  • $\begingroup$ 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 $\endgroup$ – user87543 Nov 10 '13 at 16:06
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    $\begingroup$ Yes I understand that, I just don't know how you'd go on after that $\endgroup$ – Mathlete Nov 10 '13 at 16:17
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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?

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    $\begingroup$ The first equality is because the series' converge uniformly so you can interchange the integral and summation and then the rest follows. Is that right? Is there anything more I should think about? $\endgroup$ – Mathlete Nov 10 '13 at 16:22
  • $\begingroup$ Well, I meant why the result is ""logically"" zero ? $\endgroup$ – DonAntonio Nov 10 '13 at 16:23
  • $\begingroup$ Because that is the pointwise limit of the series? $\endgroup$ – Mathlete Nov 10 '13 at 16:24
  • $\begingroup$ Nop. Look at $\;\sin nx\;$ in the interval $\;[0,2\pi]\;$ ... $\endgroup$ – DonAntonio Nov 10 '13 at 16:27
  • $\begingroup$ So it's because the area of the graph above x=0 is the same as the area below? $\endgroup$ – Mathlete Nov 10 '13 at 16:28
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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.

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