The density function of the sum of n independent uniformly distributed random variable on $(-1,1)$ is supported on $[-n,n]$ This question arose from the statement after Exercise 3.3.6 of Durrett's probability.

Exercise 3.3.6: Show that if $X_{1}, \cdots, X_{n}$ are independent uniformly distributed on $(-1,1)$, then for $n\geq 2$, $X_{1}+\cdots+X_{n}$ has density function $$f(x)=\dfrac{1}{\pi}\int_{0}^{\infty}(\sin t/t)^{n}\cos(tx)dt.$$

I proved this exercise using the inversion formula in the case of integrable characteristic function and the fact that $\sin t/t$ is an even function. 
But right after this exercise, Durrett said:

Although it is not obvious from the formula, $f$ is a polynomial in each interval $(k, k+1)$, $k\in\mathbb{Z}$, and vanishes on $[-n,n]^{c}$.

I tried to prove this statement but I did not know where to begin, since as Durrett said, it is not obvious from the formula above, then what should I do?
I may need some answer with a little bit details since I really don't know what to do now.
Thank you!
 A: The idea is to show that $f$ is the rectangle function convolved with itself, and then to show that the convolution can be expressed as a polynomial on each $(k,k+1)$.
Let $\operatorname{sinc}x = { \sin x \over x}$, and use
$\hat{g}(\omega) =  \int_{-\infty}^\infty g(t) e^{-i\omega t} dt$ to denote the Fourier transform of an $L^1$ function $g$. If $\hat{g}$ is also $L^1$, the inversion theorem (Rudin, "Real and complex analysis", Theorem 9.11, note his measure normalisation to symmetrise the formula) gives
$g(t) = {1 \over 2 \pi} \int_{-\infty}^\infty \hat{g}(\omega) e^{i\omega t} d\omega$
(interpreted in an ae. sense).
Let $r = 1_{[-1,1]}$, the symmetric rectangle function. It is straightforward to see that $\hat{r} = 2 \operatorname{sinc}$.
Note that since $\operatorname{sinc}$ is even, $f(t) = {1 \over 2 \pi} \int_{-\infty}^\infty ( \operatorname{sinc} \omega)^n e^{i\omega t} d\omega = {1 \over 2^{n+1} \pi} \int_{-\infty}^\infty \hat{r}^n(\omega) e^{i\omega t} d\omega$, which suggests that the Fourier transform (or its inverse) and convolution might be a productive approach.
In particular, it suggests that ${1 \over 2 \pi} \int_{-\infty}^\infty \hat{r}^n(\omega) e^{i\omega t} d\omega = (\underbrace{r * r *\cdots * r }_{n\text{ times}})(t)$.
A catch is that $\operatorname{sinc}$ is not an $L^1$ function so the classic inversion theorem does not apply. (Note that this is only an issue for $n=1$, but I need the $n=1$ case for the next step.)
However, since $\operatorname{sinc}$ is $L^2$ we can use the Plancherel transform (denote by ${\cal P}$). If $a,b$ are $L^2$ and $a * b$ is $L^2$ then the Parseval formula can be used to show that ${\cal P}(a * b) = {\cal P}(a) \cdot {\cal P}(b)$.
Since $r$, $r * r$, etc., are $L^2$ and $\hat{r} = 2 \operatorname{sinc}$,
we have ${\cal P}(r) = 2 \operatorname{sinc}$, and so
${\cal P} (r * r * \cdots * r) = \hat{r}^n = 2^n \operatorname{sinc}^n$.
In particular, $f(t) = {1 \over 2^n} (r * r * \cdots * r))(t)$. Note that the
equality is in an $L^2$ sense.
For $n \ge 2$, all
functions involved are $L^1 \cap L^2$, so the Plancherel transform and the usual Fourier transform are the same, and since $f$ and $r * r * \cdots * r$ are both continuous, we have $f(t) = {1 \over 2^n} (r * r * \cdots * r)(t)$ everywhere.
(Thanks to Ted Shifrin who helped clarify some confusion I was having regarding convolution with the Plancherel transform. See p.110 of Fourier Transforms by Bochner & Chandrasekharan.)
Finally, note that $ r$ is constant (hence a polynomial) on each $(k,k+1)$. So, suppose that $g = r * r * \cdots * r$ is expressible as a polynomial on each $(k,k+1)$, then
\begin{eqnarray}
(r * g)(t) &=& \int_{-\infty}^\infty r(t-s)g(s) ds \\
&=& \int_{t-1}^{t+1} g(s)ds \\
&=& \int_{t-1}^k g(s)ds + \int_k^{k+1} g(s)ds + \int_{k+1}^{t+1} g(s)ds
\end{eqnarray}
Since $g$ can be expressed as a polynomial on $(k-1,k)$ and $(k+1,k+2)$, it follows
that $r*g$ is expressible as a polynomial on $(k,k+1)$.
