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I am pretty new to calculus, and I am trying to understand some basic rules to solve Fourier Series. I don't know how I deal with the region where a function is defined and its period. For instance:

$f(x)=x^{2}$ , on $0<x<\pi$

where $f(x)$ is an odd function and periodic with a period of $4\pi$. To find out its Fourier Sine Series, I know that I need to calculate the coefficient $B_{n}$ defined as:

$B_{n}=\frac{2}{T}\int_{\lambda}^{\lambda+T}f(t)sin(\frac{nt2\pi}{T})dt$

Where $T$ is the period and $\lambda$ is any $\lambda \epsilon \mathbb{R}$. What is the interval that I should integrate? Over the region $0$ and $\pi$? Over some interval with period of $4\pi$?

Thanks,

Uirá

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  • $\begingroup$ Since $f$ is odd, we know its value on $(-\pi,\pi)$, but what is its value on $[-2\pi, -\pi] \cup [\pi, 2\pi]$? $\endgroup$ – Eric Towers Sep 8 '15 at 3:32
  • $\begingroup$ Hi Eric . The function is not defined on these intervals. As it is defined just over $0$ and $\pi$, should I integrate just over this interval? $\endgroup$ – ucaiado Sep 8 '15 at 4:17
  • $\begingroup$ Since $f$ is odd, you know what the values of $f$ on $(-\pi,0)$ are. Also, since $f$ is odd, you know $f(0)$ is zero. However, without values to "fill out" the period, the integral is also undefined. $\endgroup$ – Eric Towers Sep 8 '15 at 4:19
  • $\begingroup$ Thanks @Eric Towers, I guess that you clarified my doubt. Supposing that I had the function defined over all needed intervals, I would have to integrate over each interval, and the sum of all these intervals would result in the period desired. In this case, that the function is defined over $0<x<\pi$, if the period desired was $2\pi$, I would need to integrate over $[-\pi, \pi]$. Am I right? $\endgroup$ – ucaiado Sep 8 '15 at 4:39
  • $\begingroup$ That would be the easiest thing to do. Generally, though, sometimes functions are easier to integrate if you shift them over. As a toy example "$x+1$" is easier to integrate if you shift over by $1$ first. After you write your integral down, ask yourself if it would be easier to shift the whole thing. Also, the intervals I gave are not "magical" or "correct". Since the function is periodic, we could use any contiguous $4\pi$ interval. (We could even use chunks that aren't contiguous, as long as their periodic translates cover an entire period exactly once.) $\endgroup$ – Eric Towers Sep 8 '15 at 4:47
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I've posted a comment asking for a completion of the definition of the function $f$. However, your question about $\lambda$ can be answered generally without this information.

For the choice of $\lambda$ use whatever value makes the integral easy to evaluate. (Frequently, this means that any choice is as good as any other. This is because $\lambda$ doesn't appear in the integrand, so is only applied at the end via the Fundamental Theorem of Calculus.) Since the function is periodic, you may "stamp out" copies of the function between one choice of $\lambda$ and another. (You may need to saw off one end of the period and attach it to the other end, if the two choices of $\lambda$ don't differ by a whole period. But this won't change the value of the integral just the order that you go through the values of the integrand.) It is a good exercise to draw a picture of a periodic function with a few features per period and convince yourself that slightly increasing $\lambda$ is the same as slicing a little of the function from the left (end of $[\lambda, \lambda+T]$) and gluing it onto the right (end of that interval).

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  • $\begingroup$ Hi Eric . The function is not defined on these intervals. As it is defined just over $0$ and $\pi$, should I integrate just over this interval? $\endgroup$ – ucaiado Sep 8 '15 at 4:07
  • $\begingroup$ @ucaiado : It will help future readers if your reply to my comment is made in the same place (in the comments immediately under your question). $\endgroup$ – Eric Towers Sep 8 '15 at 4:14

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