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I have a function $f(x) = \frac{x}{\pi} \in (-\pi , \pi]$
I googled but couldn't find a solution done using complex exponential and I tired to do it as follows.

$$a_k = \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{x}{\pi}e^{-j\omega_0kx}dx$$ $$a_k = \frac{1}{2\pi ^2}\int_{-\pi}^{\pi} xe^{-j\omega_0kx}dx$$

take $$x = u\\e^{-j\omega_0kx}=\frac{dv}{dx}$$ Then by integration by parts, $$a_k = \frac{1}{2\pi ^2}\left[\left[x.\frac{e^{-j\omega_0kx}}{-j\omega_0k}\right]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}(-j\omega_0k) e^{-j\omega_0kx}dx\right]$$ $$a_k = \frac{1}{2\pi ^2}\left[\left[x.\frac{e^{-j\omega_0kx}}{-j\omega_0k}\right]_{-\pi}^{\pi}+\left[(j\omega_0k) \frac{e^{-j\omega_0kx}}{-j\omega_0k}\right]_{-\pi}^{\pi}\right]$$ $$a_k = \frac{1}{2\pi ^2}\left[\frac{\pi}{-j\omega_0k}\left(e^{-j\omega_0k \pi} + e^{j\omega_0k \pi} \right)-\left(e^{-j\omega_0k \pi} - e^{j\omega_0k \pi} \right)\right]$$ $$a_k = \frac{1}{2\pi ^2}\left[\frac{\pi}{-j\omega_0k}\cos(k\omega_0\pi)+j\sin(k\omega_0\pi)\right]$$

Can anyone give me a tip to simplify it further? Thanks!

Padmal
  • 595

1 Answers1

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There is no need of the complex exponential.

$f(x)=\frac{x}{\pi}$ is an odd function, hence we just need to compute:

$$ c_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx = \frac{2}{\pi^2}\int_{0}^{\pi} x \sin(nx)\,dx $$ and by integration by parts: $$ c_n = \frac{2}{\pi^2}\left(\left.-\frac{x}{n}\cos(nx)\right|_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\cos(nx)\,dx\right)=\frac{2(-1)^{n+1}}{\pi n}$$ from which: $$ f(x) = \sum_{n\geq 1} c_n \sin(nx) = \frac{2}{\pi}\sum_{n\geq 1}(-1)^{n+1}\frac{\sin(nx)}{n}.$$

Jack D'Aurizio
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