Proving $\int^{\infty}_0 \cos(tx)\left (\frac{\sin(t)}{t} \right )^n \, dt = 0$ I've been asked to prove that
$$ \int^\infty_0 \cos(tx)\left (\frac{\sin(t)}{t} \right )^n \, dt = 0, \space \forall x > n \geq 2.$$
My approach so far has been to use a theorem proved in class that, for a random variable $X$ with characteristic function $\phi(t)$ and $a,b\in\mathbb{R}$,
$$ \mathbb{P}(a<X<b) + \frac{\mathbb{P}(X=a) + \mathbb{P}(X=b)}{2} = \lim_{T \to \infty} \frac{1}{2\pi}\int^{T}_{-T}\frac{e^{-ita}+e^{-itb}}{it}\phi(t)\,dt.$$
So, choosing $a = -b$, I get
$$\mathbb{P}(|X| \leq b) =  \lim_{T \to \infty} \frac{1}{\pi}\int^{T}_{-T}\frac{\sin(t)}{t}\phi(t)\,dt$$
which makes it seem like I need to find a random variable that satisfies $\phi(t)$ and $\mathbb{P}(|X|<b)=0$... but I'm losing confidence in this approach, since it doesn't account for why $x > n \geq 2$ is required or incorporate the behavior of $\left(\frac{\sin t}{t}\right)^n$.
A solution using probability techniques would be preferable to pure analysis, but all help is appreciated.
 A: 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.
A: A complex analysis approach is to integrate the function $e^{ixz} \left(\frac{\sin z}{z} \right)^{n} $ around a contour that consists of the real axis and the large semicircle above it.
To show that the integral along the semicircle vanishes as the radius of the semicircle goes to $\infty$, we only need to show that $e^{ixz} \sin^{n}(z)$ is bounded in the the upper half-plane.
But this is clearly the case since if $x \ge n$, then $$e^{ixz} \sin^{n}(z) = e^{ixx} \left(\frac{e^{ix}-e^{-ix}}{2i} \right)^{n}= \frac{e^{ixz}}{(2i)^{n}} \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}e^{ikz}e^{-i(n-k)z}$$ is a linear combination of exponential functions of the form $e^{iaz}$, where $a \ge 0$. And in the upper half-plane, the magnitude of $e^{iaz}, \ a \ge 0$, is less than or equal to $1$.
Therefore, since the singularity at the origin is removable,  we get $$\int_{-\infty}^{\infty} e^{ixt} \left(\frac{\sin t}{t} \right)^{n} \, dt =0,$$ which leads to the result if we equate the real parts on both sides of the equation.
