FS of ±1.5 square wave, ω₀ = π

The fundamental frequency is $\omega_0 = \pi$, so the period is $T = 2\pi/\omega_0 = 2$. Over one period, $x(t) = 1.5$ on $0 \le t < 1$ and $x(t) = -1.5$ on $1 \le t < 2$.

 x(t)
  1.5  ____      ____      ____
      |    |    |    |    |
 -----+----+----+----+----+----- t
      0    1    2    3    4
           |____|    |____|
 -1.5

DC coefficient

$$a_0 = \frac{1}{T}\int_0^2 x(t)\,dt = \frac{1}{2}\left[1.5(1) - 1.5(1)\right] = 0$$

Coefficients for k ≠ 0

$$a_k = \frac{1}{T}\int_0^T x(t)e^{-jk\omega_0 t}dt = \frac{1}{2}\left[\int_0^1 1.5\,e^{-jk\pi t}dt - \int_1^2 1.5\,e^{-jk\pi t}dt\right]$$

Evaluate each integral, using $e^{-jk\pi} = (-1)^k$ and $e^{-j2k\pi} = 1$:

$$\begin{aligned} \int_0^1 e^{-jk\pi t}dt &= \frac{1 - e^{-jk\pi}}{jk\pi} = \frac{1-(-1)^k}{jk\pi} \\ \int_1^2 e^{-jk\pi t}dt &= \frac{e^{-jk\pi} - e^{-j2k\pi}}{jk\pi} = \frac{(-1)^k - 1}{jk\pi} \end{aligned}$$

Substituting:

$$\begin{aligned} a_k &= \frac{1.5}{2}\cdot\frac{[1-(-1)^k] - [(-1)^k - 1]}{jk\pi} \\ &= \frac{1.5}{2}\cdot\frac{2[1-(-1)^k]}{jk\pi} = \frac{1.5\,[1-(-1)^k]}{jk\pi} \end{aligned}$$

So

$$a_k = \begin{cases} \dfrac{3}{jk\pi} = -j\dfrac{3}{k\pi}, & k \text{ odd} \\ 0, & k \text{ even} \end{cases}$$

The coefficients are purely imaginary and odd ($a_{-k} = -a_k$). This is expected, because $x(t)$ is real and has odd symmetry about $t = 0$ (half-wave symmetry also removes the even harmonics). In trigonometric form:

$$x(t) = \frac{6}{\pi}\left[\sin\pi t + \frac{1}{3}\sin 3\pi t + \frac{1}{5}\sin 5\pi t + \dots\right]$$

Answer: $a_0 = 0$, $a_k = -j\,3/(k\pi)$ for odd $k$, and $a_k = 0$ for even $k$.