Fourier coefficients of odd and even part of a signal
OK, so your signal is described in one period, $T_0$, as $$x_{T_0}(t) = \left\{\begin{array}{ll}
0, &amp; -T_0/2 &lt; t \leq 0 \\
1, &amp; 0 &lt; t \leq T_0/2
\end{array}\right.$$
Let's keep it that way (without any other shortcuts like $\prod$ or $\mathrm{rect(\cdot)}$ etc).
One way to find the Fourier coefficients of the even and the odd part of a signal is to actually find the even and odd part of the signal, and do a Fourier Series analysis on each. 

But, another way is to check Fourier Series properties - again :-) . 

If $x(t)$ is real - which it is - with Fourier coefficients $X_k$, then the odd part of $x(t)$, let it be $x_{odd}(t)$, has Fourier coefficients $X^{odd}_k = j\Im\{X_k\}$, where $\Im\{X_k\}$ is the imaginary part of the Fourier coefficients of the original signal $x(t)$. 

Similarly, the even part of the signal $x(t)$, let it be $x_{ev}(t)$, has Fourier coefficients $X^{ev}_k = \Re\{X_k\}$, where $\Re\{X_k\}$ is the real part of the original Fourier coefficients $X_k$.

In short: $$x(t) \longleftrightarrow X_k$$ $$x_{odd}(t) \longleftrightarrow j\Im\{X_k\}$$ $$x_{even}(t) \longleftrightarrow \Re\{X_k\}$$

So, you have two options:


Find $x_{odd}(t)$, $x_{ev}(t)$ and find their Fourier coefficients analytically (takes a lot of time and paper)
Find the Fourier coefficients, $X_k$, of $x(t)$ and then take their real and imaginary part (faster and easier)


I would go for the second option, since $x(t)$ is very simple and its Fourier coefficients are given by $$X_k = \frac{1}{T_0}\int_0^{T_0}x_{T_0}(t)e^{-j2\pi kf_0t}dt = \frac{1}{T_0}\int_0^{T_0/2}1\cdot e^{-j2\pi kf_0t}dt$$

P.S: In case you wish to go with the first option, the even part of the signal can be computed by first finding $$x_{T_0}(-t) = \left\{\begin{array}{ll}
1, &amp; -T_0/2 \leq t \leq 0 \\
0, &amp; -T_0 &lt; t &lt; -T_0/2
\end{array}\right.$$
and then $$x_{even, T_0}(t) = \frac{1}{2}(x_{T_0}(t) + x_{T_0}(-t)) = \frac{1}{2}$$

Similarly, the odd part can be obtained as $$x_{odd,T_0}(t) = \frac{1}{2}(x_{T_0}(t)-x_{T_0}(-t)) = \left\{\begin{array}{ll}
-\frac{1}{2}, &amp; -T_0/2 &lt;t &lt; 0 \\
\frac{1}{2}, &amp; 0 \leq t \leq T_0/2
\end{array}\right.$$

So the even part is just the mean value of the signal, $X_0$. The odd part of the signal has Fourier coefficients $$X_k = -\frac{1}{T_0}\int_{-T_0/2}^0 \frac{1}{2}e^{-j2\pi kf_0t}dt + \frac{1}{T_0}\int_0^{T_0/2}\frac{1}{2}e^{-j2\pi kf_0t}dt$$

I think you can continue from here. Let me know if you make it till the end. :)