Fourier series coefficients and plot

The paper does not print the signal. This answer takes the standard textbook case: a periodic square wave of period $T$ with $x(t) = 1$ for $|t| < T_1$ and $x(t) = 0$ for $T_1 < |t| < T/2$.

 x(t)
  1     ___       ___       ___
       |   |     |   |     |   |
 ______|   |_____|   |_____|   |___ t
     -T1   T1  T-T1  T+T1
       (pulse centred at t = 0)

Coefficients

With $\omega_0 = 2\pi/T$, the DC term is

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

For $k \ne 0$:

$$\begin{aligned} a_k &= \frac{1}{T}\int_{-T_1}^{T_1} e^{-jk\omega_0 t}dt = \frac{1}{T}\left[\frac{e^{-jk\omega_0 t}}{-jk\omega_0}\right]_{-T_1}^{T_1} \\ &= \frac{2}{k\omega_0 T}\cdot\frac{e^{jk\omega_0 T_1} - e^{-jk\omega_0 T_1}}{2j} = \frac{\sin(k\omega_0 T_1)}{k\pi} \end{aligned}$$

The coefficients are real, because $x(t)$ is real and even.

Example values for T = 4T₁ (50% duty cycle)

Here $\omega_0 T_1 = \pi/2$, so $a_k = \frac{\sin(k\pi/2)}{k\pi}$:

Plot of the coefficients

 a_k
  0.50                   |
  0.32                |  |  |
  0.06    |           |  |  |           |
  0  -----+--+--+--+--+--+--+--+--+--+--+--- k
         -5 -4 -3 -2 -1  0  1  2  3  4  5
 -0.11          |                 |

The coefficients are samples, at $\omega = k\omega_0$, of the sinc-shaped envelope $\frac{\omega_0}{\pi}\cdot\frac{\sin(\omega T_1)}{\omega}$. Even harmonics vanish for a 50% duty cycle, and the odd ones decrease as $1/k$ with alternating sign.