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The question is to find the complex Fourier series expansion of

$$ f(x) = x , -\pi < x < \pi $$ I solved for $C_n$ and got:

$$ C_n = \frac{1}{n}(-1)^n - \frac{1}{\pi n^2 } \sin(n \pi ) $$

But I saw a solution online and saw that they got:

$$ C_n = \frac{i}{n}(-1)^n - \frac{i}{\pi n^2 } \sin(n \pi ) $$

$$ f(x) = \sum_{n = -\infty}^{\infty} \left( \frac{i}{n}(-1)^n - \frac{i}{\pi n^2 } \sin(n \pi )\right)e^{inx} $$

Can someone tell me why the numerator has an i? And also since $\sin(n \pi ) = 0$, would the answer then be:

$$ f(x) = \sum_{n = -\infty}^{\infty} \left( \frac{i}{n}(-1)^n \right)e^{inx} $$

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$$c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}xe^{-inx}dx=\Big[\frac{i}{2n\pi}xe^{-inx}\Big]_{-\pi}^{\pi}-\frac{i}{2n\pi}\int_{-\pi}^{\pi}e^{-inx}dx=$$ $$=\frac{i}{n}\cos(\pi{n})+\frac{1}{2\pi{n}^{2}}\Big[e^{-inx}\Big]_{-\pi}^{\pi}=$$ $$=\frac{i}{n}\cos(\pi{n})-\frac{i}{\pi{n}^{2}}\sin(\pi{n})=\frac{i}{n}(-1)^{n}$$

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  • $\begingroup$ how did you get the i in the numerator? @KirylPesotski $\endgroup$ – gofish Feb 26 '17 at 20:38
  • $\begingroup$ in comes out from the integration by parts $\endgroup$ – Kiryl Pesotski Feb 27 '17 at 13:41

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