Spectrum of a sawtooth wave

One period: $x(t) = t$ for $0 \le t < 1$, with period $T = 1$ s, so $\omega_0 = 2\pi/T = 2\pi$ rad/s.

$$a_k = \frac{1}{T}\int_0^T x(t)e^{-jk\omega_0 t}dt = \int_0^1 t\,e^{-j2\pi kt}dt$$

DC term ($k = 0$)

$$a_0 = \int_0^1 t\,dt = \frac{1}{2}$$

Coefficients for $k \ne 0$

Integrate by parts with $u = t$, $dv = e^{-j2\pi kt}dt$, $v = \frac{e^{-j2\pi kt}}{-j2\pi k}$:

$$\begin{aligned} a_k &= \left[\frac{t\,e^{-j2\pi kt}}{-j2\pi k}\right]_0^1 - \int_0^1\frac{e^{-j2\pi kt}}{-j2\pi k}dt \\ &= \frac{e^{-j2\pi k}}{-j2\pi k} + \frac{1}{j2\pi k}\left[\frac{e^{-j2\pi kt}}{-j2\pi k}\right]_0^1 \\ &= \frac{1}{-j2\pi k} + \frac{1}{j2\pi k}\cdot\frac{e^{-j2\pi k} - 1}{-j2\pi k} \end{aligned}$$

Since $e^{-j2\pi k} = 1$, the second term is zero:

$$a_k = \frac{-1}{j2\pi k} = \frac{j}{2\pi k}, \qquad k \ne 0$$

Magnitude and phase

$$|a_k| = \frac{1}{2\pi|k|}, \qquad \angle a_k = \begin{cases} +90^\circ\ (\pi/2), & k > 0 \\ -90^\circ\ (-\pi/2), & k < 0 \\ 0, & k = 0 \end{cases}$$

So the Fourier series is

$$x(t) = \frac{1}{2} + \sum_{k\ne0}\frac{j}{2\pi k}e^{j2\pi kt} = \frac{1}{2} - \sum_{k=1}^{\infty}\frac{1}{\pi k}\sin(2\pi kt)$$

Spectra (lines at $\omega = 2\pi k$)

 |a_k|
   0.5           |
 0.159        |  |  |
0.0796     |  |  |  |  |
0.0531  |  |  |  |  |  |  |
       -+--+--+--+--+--+--+--> k
       -3 -2 -1  0  1  2  3

 angle a_k (deg)
    90              |  |  |
       -+--+--+--o--+--+--+--> k
   -90  |  |  |
       -3 -2 -1  0  1  2  3

The magnitude falls as $1/k$ (because of the jump discontinuity each period), is even in $k$, and the phase is odd: $+90^\circ$ for positive $k$ and $-90^\circ$ for negative $k$.

Check: at $t = 0.25$, the series $\frac{1}{2} - \frac{1}{\pi}\left(1 - \frac{1}{3} + \frac{1}{5} - \dots\right) = \frac{1}{2} - \frac{1}{\pi}\cdot\frac{\pi}{4} = 0.25$, which equals $x(0.25)$.