Proving $\int^{\infty}_0 \cos(tx)\left (\frac{\sin(t)}{t} \right )^n \, dt = 0$

Hint Let $X_1,\ldots,X_n$ be independent random variables such that $X_j$ is uniformly distributed on $[-1,1]$ for $j=1,\ldots,n$. It is not difficult so show that the Fourier transform of $X_j$ equals

$$\Phi_{X_j}(t) := \mathbb{E}e^{\imath \, t \cdot X_j} = \frac{\sin t}{t}.$$

Therefore, the Fourier transform of $Y := X_1+\ldots+X_n$ is given by

$$\Phi_Y(t) = \left( \frac{\sin t}{t} \right)^n.$$

Note that

$$\begin{align*} \int_0^{\infty} \cos(t x) \cdot \left( \frac{\sin t}{t} \right)^n \, dt &= \text{Re} \left( \frac{1}{2} \int_{-\infty}^{\infty} e^{-\imath \, t \cdot x} \cdot \Phi_Y(t) \, dt \right). \end{align*}$$

Now use that the right-hand side is basically the (real part of the) inverse Fourier transform to prove the assertion.