Fourier series of translated square I can't seem to find the correct Fourier series coefficients ($s_n$) of the following periodic function. I know how to get the Fourier series of the same one that is not vertically translated and has no negative values for any $t$ (Result is $A sinc(n\pi)$), but this one I'm having trouble with. The function is periodic, and during it's period T, is defined with:
$$
  x(t) =
\begin{cases}
A,  & t\in(0,\frac{T}{4}) \\
-A, & t\in(\frac{T}{4},\frac{3T}{4}) \\
A, & t\in(\frac{3T}{4},T)
\end{cases}
$$
How I tried solving this:
$$ s_n = \frac{1}{T}\int_{-T/2}^{T/2}{x(t)e^{-inw_0t}dt} = \frac{1}{T}\Bigg[\int_{-T/2}^{-T/4}{-Ae^{-inw_0t}dt} + \int_{-T/4}^{T/4}{Ae^{-inw_0t}dt} + \int_{T/4}^{T/2}{-Ae^{-inw_0t}dt}\Bigg]$$
Integrating I get:
$$ s_n = \frac{A}{inw_0T}\Bigg[ (e^{inw_0\frac{T}{2}}-e^{-inw_0\frac{T}{2}})-2(e^{inw_0\frac{T}{4}}-e^{-inw_0\frac{T}{4}})\Bigg]$$
Applying Euler's identity and $w_0 = \frac{2\pi}{T}$ and the definition of the $sinc$ function finally I get:
$$ s_n = Asinc(n\pi) - Asinc(\frac{n\pi}{2})$$
That differs from the answer I have in my textbook which says the answer is only $A sinc(\frac{n\pi}{2})$. 
It would help if someone could shed some light on where I've done wrong.

Since it is said that $x$ is $T$-periodic, you have that
$x(t) = -A$ for $t\in(-T/2,-T/4)$, so you will have to re-evaluate the integrals.
Then, note that
$$\text{sinc}(n\pi) = 0$$
