Fourier Transforms of shifted sinc funtions I would like to calculate the Fourier transform of the following functions:


*

*$$\left(\dfrac{\sin(\pi x\pm\pi n/2)}{\pi x\pm\pi n/2}\right)^2$$

*$$\dfrac{\sin(\pi x+\pi n/2)}{\pi x+\pi n/2}\cdot\frac{\sin(\pi x-\pi n/2)}{\pi x-\pi n/2}$$
with $n \in\mathbb{N}$
Any help will be highly appreciated!
 A: Hint 1: the transform of $\sin (x\pi)/(x\pi)$ is a rectangle with height 1 between $-\pi$ and $\pi$.
Hint 2: multiplication in one domain corresponds to convolution in the other domain.
Hint 3: shifting a function by $x_0$ corresponds to multiplication of its Fourier transform with $e^{\pm ix_0\omega}$.
EDIT: For solving (2.) we have the Fourier transforms of the two functions:
$$e^{\pm i\omega n/2}\text{rect}(\omega,\pi)$$
So the convolution becomes
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\Omega n/2}\text{rect}(\Omega,\pi)
e^{-i(\omega-\Omega) n/2}\text{rect}(\omega-\Omega,\pi)d\Omega=\\
=\frac{e^{-i\omega n/2}}{2\pi}\text{rect}(\omega,2\pi)\int_{\max (-\pi,\omega-\pi)}^{\min (\pi,\omega+\pi)}e^{j\Omega n}d\Omega$$
For $-2\pi<\omega < 0$ we get
$$\frac{e^{-i\omega n/2}}{2\pi}\int_{-\pi}^{\pi+\omega}e^{j\Omega n}d\Omega$$
and for $0<\omega<2\pi$ we have
$$\frac{e^{-i\omega n/2}}{2\pi}\int_{\omega-\pi}^{\pi}e^{j\Omega n}d\Omega$$
I trust you can take it from here.
EDIT 2:
OK, some more help and the final solution. The first integral (for $-2\pi<\omega < 0$ ) evaluates to
$$\frac{e^{-i\omega n/2}}{2\pi}\frac{(-1)^n}{in}\left (e^{in\omega} -1\right )=
\frac{(-1)^n}{n\pi}\sin\frac{n\omega}{2}$$
and the second integral for positive $\omega$ is the same, just with the opposite sign. So the final result is
$$\frac{(-1)^{n+1}}{n\pi}\text{sign}(\omega)\sin\frac{n\omega}{2},\quad
|\omega|<2\pi,\quad \text{(zero otherwise)}$$
