FS coefficients of periodic square wave

The signal has period $T$, so $\omega_0 = 2\pi/T$. Over one period ($-T/2$ to $T/2$) it equals $A$ only for $|t| < T_1$.

$$a_k = \frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_{-T_1}^{T_1}A\,e^{-jk\omega_0 t}dt$$

DC term ($k = 0$)

$$a_0 = \frac{1}{T}\int_{-T_1}^{T_1}A\,dt = \frac{2AT_1}{T}$$

This is the average value (height times duty cycle).

For $k \ne 0$

$$\begin{aligned} a_k &= \frac{A}{T}\left[\frac{e^{-jk\omega_0 t}}{-jk\omega_0}\right]_{-T_1}^{T_1} = \frac{A}{T}\cdot\frac{e^{jk\omega_0 T_1} - e^{-jk\omega_0 T_1}}{jk\omega_0} \\ &= \frac{2A}{k\omega_0 T}\sin(k\omega_0 T_1) \\ &= \frac{A\sin(k\omega_0 T_1)}{k\pi} \qquad (\omega_0 T = 2\pi) \end{aligned}$$

Result

$$a_k = \begin{cases} \dfrac{2AT_1}{T}, & k = 0 \\[2mm] \dfrac{A\sin(k\omega_0 T_1)}{k\pi} = \dfrac{2AT_1}{T}\,\text{sinc}\left(\dfrac{2kT_1}{T}\right), & k \ne 0 \end{cases}$$

(with $\text{sinc}(x) = \sin(\pi x)/(\pi x)$). The coefficients are real and even because $x(t)$ is real and even.

Special case, 50% duty cycle ($T = 4T_1$): $\omega_0 T_1 = \pi/2$, so $a_0 = A/2$ and $a_k = \frac{A\sin(k\pi/2)}{k\pi}$: $a_{\pm1} = A/\pi$, $a_{\pm2} = 0$, $a_{\pm3} = -A/3\pi$, $a_{\pm5} = A/5\pi$, i.e. even harmonics are zero.

 a_k / A  (T = 4T1)
   0.5                 |
 0.318              |  |  |
0.0637  |           |  |  |           |
       -+--o--+--o--+--+--+--o--+--o--+--> k
-0.106        |                 |
       -5 -4 -3 -2 -1  0  1  2  3  4  5

The envelope of $a_k$ follows a sinc shape; making the pulse narrower (smaller $T_1/T$) spreads the spectrum wider.