The characteristic function of the uniform distribution for $t\neq 0$ is:
$$\Large\varphi(t)=\frac{e^{itu}-e^{itl}}{(u-l)it}$$
with $u$ and $l$ corresponding to the upper and lower limits of the pdf:
$$f_X(x) = \frac{1}{u-l}$$
In the Wikipedia entry for characteristic function there is this plot:
with the caption:
The characteristic function of a uniform U(–1,1) random variable. This function is real-valued because it corresponds to a random variable that is symmetric around the origin; however characteristic functions may generally be complex-valued.
However, I'm stuck with a probably silly step:
$$\varphi(t)=\frac{e^{1it}-e^{-1it}}{(1-(-1))it}=\frac{e^{0}}{2it}= \frac{1}{2it}$$
How do you get rid of the imaginary part in the denominator?
CONCLUSION:
$$\varphi(t)=\frac{e^{it}-e^{-it}}{(1-(-1))it}=\frac{2i\sin(t)}{2it}=\frac{\sin(t)}{t}$$
This is the sinc function.
It seems to me that you believe that $$e^{it} - e^{-it} = e^0$$ but this is wrong. Instead you have to use the identity $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}.$$