Fourier series expansion of $x(t) = \sum\nolimits_{z \in \mathbb{Z}} (-1)^z \delta(t - 2z)$ 
Find the Fourier series expansion of $x(t) = \sum\nolimits_{z \in \mathbb{Z}} (-1)^z \delta(t - 2z)$, where $\delta(\cdot)$ denotes the Dirac delta function (unit impulse).

I can infer that the period of $x(t)$ is $T = 4$.
However, I am stuck in finding the Fourier coefficients
$$
C_k = \frac1T \int\limits_{\tau_0}^{\tau_0 + T} x(\tau) e^{-i\frac{2 \pi k \tau}{T}} \operatorname d \tau
$$
where $\tau_0 \in \mathbb{R}$, $T = 4$ is the period, and $x(t)$ is defined as above. Can anyone give me some help as to how to handle that integral?
 A: The function $x(t)$ given by
$$x(t)=\sum_{k=-\infty}^{\infty} (-1)^k\delta(t-2k)$$
is periodic with period $4$. 
We can write the Fourier Series for $x(t)$ as
$$x(t)=\sum_{n=-\infty}^{\infty}c_ne^{i\pi nt/2}$$ 
where the Fourier coefficients $c_n$ are given by
$$\begin{align}
c_n&=\frac14\int_0^4 x(t)e^{-i\pi nt/2}dt\\\\
&\frac14 \int_0^4 \left(\delta(t)-\delta(t-2)\right)e^{-i\pi nt/2}dt\\\\
&=\frac14\left(1-e^{-i \pi n}\right)\\\\
&=\frac{1-(-1)^n}{4}
\end{align}$$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{x(t)\sim \sum_{n=-\infty}^{\infty}\frac{1-(-1)^n}{4}e^{i\pi nt/2}=\sum_{n=1}^{\infty}\cos\left(\frac{(2n-1)\pi t}{2}\right)}$$
It is interesting to note that by formal term-by=term integration of the Fourier Series representation of $x(t)$, we have for $t\in [0,4)$
$$\int_0^t x(t')dt'=\sum_{n=1}^{\infty}\frac{\sin\left(\frac{(2n-1)\pi t}{2}\right)}{\frac{(2n-1)\pi }{2}}$$
which we recognize as the Fourier Series for the pulse function $p(t)$ given by
$$
p(t)=
\begin{cases}
1&,t\in [0,2)\\\\
\frac12&,t=2\\\\
0&,t\in(2,4)
\end{cases}
$$
