Evaluate $\hat f(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-itx}dt$ for $f(t)=\sum_{n=0}^{\infty}\frac{1}{2^n}1_{[-2,2]}(x-2n)$ 
$\hat f(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-itx}dt$, for $f \in L^1(\mathbb R)$
Let $f(t)=\sum_{n=0}^{\infty}\frac{1}{2^n}1_{[-2,2]}(t-2n)$, where $1_{[.,.]}$ is the indicator function
I want to determine $\hat f(x).$

I think we need to interchange the series and the integral which should be no problem because
$\frac{1}{2^n}1_{[-2,2]}(x-2n) \ge 0$ so we can use Fubini.
Now, using $1_{[-2,2]}(x-2n)=1_{[2n-2,2n+2]}(x)$:
$\hat f(x)=\frac{1}{\sqrt{2\pi}}\sum_{n=0}^{\infty}\frac{1}{2^n}\int_{-\infty}^{\infty}1_{[2n-2,2n+2]}(t)e^{-itx}dt=\frac{1}{\sqrt{2\pi}}\sum_{n=0}^{\infty}\frac{1}{2^n}\frac{e^{-i(2n+2)}-e^{-i(2n-2)}}{-ix}$
Is that correct so far?
I thought about using $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$ to simplify this term but I don't know if that will help

If you are also required to be rigorous you have to justify the interchange of the integral and the infinite sum. You can do this by observing that  your sereis converges in $L^{1}$ norm and $s_n \to f$ in $L^{1}$ norm  implies $\hat {s_n} \to \hat {f}$ pointwise. [Take $s_n$ to be the n-th partial sum of the series]. 
