Show that $S_n = X_1+ \cdots X_n $ the probability density function $f_{s{_n}}$ We consider $(\Omega, \mathcal{F}, \mathcal{P})$ a probability space and $(X_{n})_{n \in \mathbb{N}}$ a sequence of independent and identically distributed random variables such that $X_1 \sim \operatorname{Unif}(-1,1)$. Show that $S_n = X_1+ \cdots X_n $ for $n \ge 2$ has  the probability density function $$f_{s{_n}} = \frac{1}{n} \int_{0}^{\infty} \Bigg(\frac{\sin(t)}{t} \bigg)^n \cos(xt)dt $$
I have found out that I could use that $|\frac{\sin(t)}{t}|\le1$ for $\forall t \in \mathbb{R}$ and I have also calculated that $$\int_{0}^{\infty} \Bigg(\frac{\sin(t)}{t} \bigg)^2 dt = \frac{\pi}{2}$$ At this point I have no clue how should I continue it I would really appreciate any kind of help since I'm learning the topics on my own.

The characteristic function of $S_n$ is
$$\phi_{S_n}(t) = E[e^{itS_n}] = \prod_{k=1}^n E[e^{itX_k}] = \left(\frac{e^{it} - e^{-it}}{2it}\right)^n = \left(\frac{\sin t}{t}\right)^n.$$
The inversion theorem shows how you can compute the density $f_{S_n}$ if you know the characteristic function $\phi_{S_n}$.
\begin{align}
f_{S_n}(x)
&= \frac{1}{2\pi} \int_{-\infty}^\infty e^{-itx} \phi_{S_n}(t) \, dt
\\
&= \frac{1}{2\pi} \int_0^\infty e^{-itx} \phi_{S_n}(t) \, dt + \frac{1}{2\pi} \int_0^\infty e^{iux} \phi_{S_n}(-u) \, du
& u = -t
\\
&= \frac{1}{2\pi} \int_0^\infty (e^{itx} + e^{-itx}) \phi_{S_n}(t) \, dt
& \phi_{S_n}(-t) = \phi_{S_n}(t)
\\
&= \frac{1}{\pi} \int_0^\infty \cos(tx) \phi_{S_n}(t) \, dt.
\end{align}
