I am having difficulties to calculate the fourier transform of $(\frac{\sin(t)}{t})^{6}$.
It could be rewritten as $(-15 \cos(2 t) + 6 \cos(4 t) - \cos(6 t) + 10)/(32 t^6)$.
In this case, I still have to deal with the $(32t^6)^{-1}$. Can someone please help me out?
Hint: use the convolution theorem. This will let you write
$$ \mathcal{F}[f(t)g(t)] = \hat{f}(\omega)*\hat{g}(\omega) $$ Where $*$ is the convolution. Similarly,
$$ \mathcal{F}[(f(t))^n] = *^{(n)}\hat{f}(\omega) $$ where $*^{(n)}$ is the n-fold convolution of $\hat{f}$ with itself. Since the Fourier transform of $\sin(t)/t$ is easy to compute (or look up in a table), you should be able to proceed from here.