Sawtooth wave Fourier coefficients
This is one of the best problems to demonstrate Fourier Series properties, and specifically the time derivative property: $$\frac{d}{dt}x(t) \overset{FS}\longleftrightarrow j2\pi kf_0 X_k$$

Instead of computing the integral $$X_k=\frac{1}{T_0}\int_{T_0}x(t)e^{-j2\pi kf_0t}dt$$ which is time consuming, you can take the first derivative of your signal and compute the Fourier coefficients of the derivative. At discontinuities, you will have Delta functions, and everywhere else you will have a constant equal to the slope of your $y$ signal, which is $2A/T_0$. In other words, inside the time interval $[0,T_0)$, your derivative can be expressed as $$\frac{d}{dt}x_{T_0}(t) = \Bigg\{\begin{array}{ll}\frac{2A}{T_0}, &amp; t \in (0,T_0) \\ -2A, &amp; t=0\end{array} = \frac{2A}{T_0}\mathrm{rect}\Big(\frac{t-\frac{T_0}{2}}{T_0}\Big)-2A\delta(t)$$
I will not use the last equality for integration, but you now see it is much easier to integrate $\frac{d}{dt}x_{T_0}(t)$. Let's say $z(t)=\frac{d}{dt}x(t)$, so its Fourier coefficients would be $$\begin{align}
Z_k &amp;= \frac{1}{T_0}\int_0^{T_0}\frac{2A}{T_0}e^{-j2\pi kf_0t}dt -\frac{2A}{T_0}\int_{0}^{T_0}\delta(t)e^{-j2\pi kf_0t}dt \\
&amp;= \frac{2A}{T_0^2}\frac{1}{(-j2\pi kf_0)}e^{-j2\pi kf_0t}\Big|_{0}^{T_0} - \frac{2A}{T_0}e^{-j2\pi kf_0t}\Big|_{t=0} \\
&amp;= \frac{2A}{-j2\pi kf_0T_0^2}(e^{-j2\pi kf_0T_0}-1)-\frac{2A}{T_0} \\
&amp;= 0 - \frac{2A}{T_0} = -\frac{2A}{T_0}
\end{align}$$
Can you apply the time derivative property to find your $X_k$ coefficients?

P.S: If you are not aware of the Delta function and its properties, or even the properties of Fourier Series, then just skip my answer. :)