Find $0$-th Fourier series coefficient of the $3\pi$ periodic function $h(t)$. What went wrong?

Question

I'm asked to find the $0$-th coefficient of the Fourier series representation of the $3\pi$ periodic function $h(t)$

I proceed as follows: $$\frac{1}{3\pi} \int_{0}^{3\pi} \Bigl(\frac{t}{\pi}\Bigr) + \Bigl(\frac{t}{\pi}\Bigr)\, dt = \frac{2}{3\pi}\int_{0}^{3\pi}\Bigl(\frac{t}{\pi}\Bigr)\, dt = ... = 3$$

But the solution given to me says otherwise: $$\frac{1}{3\pi} \int_{-3\pi/2}^{3\pi/2} h(t)\, dt = \frac{2}{3\pi}\int_{0}^{\pi}\frac{t}{\pi}\, dt = \cdots = \frac{1}{3}$$

May I ask you what I forget to take into account? Currently, I have some troubles finding out why. Thanks a lot !

Answer 1

On $[0,3\pi$], the periodic function $h$ is given by $$h(t)= \begin{cases} \tfrac{t}{\pi} & 0\leq t< \pi\\ 0 & \pi\leq t< 2\pi\\ -\tfrac{t-3\pi}{\pi} & 2\pi\leq t< 3\pi \end{cases}$$ Then $$a_0=\frac{1}{3\pi}\int_{0}^{3\pi}h(t)dt=\frac{1}{3\pi}\int_0^{\pi}\frac{t}{\pi}dt+\frac{1}{3\pi}\int_{2\pi}^{3\pi} \Big[-\frac{t-3\pi}{\pi}\Big]dt$$ By symmetry $$\frac{1}{3\pi}\int_{2\pi}^{3\pi} \Big[-\frac{t-3\pi}{\pi}\Big]dt=\frac{1}{3\pi}\int_0^{\pi}\frac{t}{\pi}dt$$ thus $$a_0=2\cdot \frac{1}{3\pi}\int_0^{\pi}\frac{t}{\pi}dt=\frac{2}{3\pi}\cdot \frac{\pi}{2}=\frac{1}{3}$$

Answer 2

$$a_0=\frac{1}{6\pi}\int_{-3\pi}^{3\pi} h(t)\mathrm{d}t$$ The easiest way to do this is just add up the areas of the four triangles. There are four of them on the $[-3\pi,3\pi]$ interval each with a base of $\pi$ and height of $1$ so - $$a_0=\frac{1}{6\pi}\left(4\cdot \frac{1\cdot \pi}{2}\right)$$ $$a_0=\frac{1}{3}$$