1
$\begingroup$

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

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

  • 2
    How about $\int_0^\pi dx e^{ikx}$?2011-03-07

1 Answers 1

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...