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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2How about $\int_0^\pi dx e^{ikx}$? – Fabian Mar 07 '11 at 09:33
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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...
Thomas Rot
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