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I was practicing Fourier Series earlier for an exam I have coming up, and I noticed something that has got me incredibly confused. I'll use the following problem to demonstrate: Periodic Function.

The function is evidently odd, so it will contain no cosine terms or constants. The problem I have is working out the $b_n$ term associated with the sine terms:

If I simply integrate between $t_0$ and $-t_0$, I will achieve a value of zero, and thus all terms in the Fourier Series will be zero, which does not make sense. In the answer, I see that they choose to integrate between $t_0$ and $0$ and then multiply by two, which achieves a totally different answer.

My question is how do I know when to do this? For every other function I have come across, it has always been sufficient to simply integrate over the period, rather than integrating across half of the period and then multiplying by 2, as they do here: Solution.

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  • $\begingroup$ If $f$ is odd, then $t \mapsto f(t)\cdot\sin (\lambda t)$ is even, so the integral over $[-t_0,t_0]$ need not vanish. $\endgroup$ – Daniel Fischer Jan 12 '17 at 11:23
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You have done something wrong. What, I don't know, since you have not shown your computations !

I get: $\int_{-t_0}^{t_0}F(t) \sin (nt)dt=\frac{2F_0}{n}(1- \cos(n t_0))$

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  • $\begingroup$ Ah I see my mistake now, thank you. Are they able to rewrite the integral as they did because the function being integrated is even? If it were an odd function, they would surely not be able to rewrite the integral in this way. $\endgroup$ – user406315 Jan 12 '17 at 11:33

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