# Fourier Transform of $(\frac{\sin(t)}{t})^{6}$

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?

$$\mathcal{F}[f(t)g(t)] = \hat{f}(\omega)*\hat{g}(\omega)$$ 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.
• The answer will be $1/2\pi$ times the probability density function of the sum of six independent and identically-distributed random variables, each uniformly distributed between -1 and 1. Thus the support of the result will be $-6<\omega<+6$. It won't be a very nice-looking expression, I'm afraid, and will be a rather complicated piece-wise function. – John Barber Dec 4 '16 at 6:18
• @Anni_housie yes, it will be band-limited. The Fourier transform of $\sin(t)/t$ is a `box' function, so the convolution of the box with itself 6 times will still be band-limited. It'll be a 6th order spline function (piecewise polynomial), so not pretty! – icurays1 Dec 4 '16 at 6:21
• By the central limit theorem, the answer will be close to a normal distribution in $\omega$, with mean 0 and standard deviation $\sqrt{2}$. – John Barber Dec 4 '16 at 7:02