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How can I compute of the $2\pi$-periodic function whose restriction to the range $[-\pi,\pi]$ is:

$$ \begin{cases} 1 && 0 < x < \pi \\ 0 && -\pi < x < 0 \end{cases} $$

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How about $\int_0^\pi dx e^{ikx}$? –  Fabian Mar 7 '11 at 9:33

1 Answer 1

By working out the integral $c_n=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-i n x}dx$, which in this case simplifies to

$$c_n=\frac{1}{2\pi}\int_0^{\pi}e^{-i n x}dx$$

From here on it shouldn't be too hard...

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