Proving that $\int_1^\infty \big(\frac{\sin x}{x}\big)^n d x\leq e^{-n/6}$ How to prove that, for $n\geq 1$
$$\int_1^\infty \big(\frac{\sin x}{x}\big)^n d x\leq e^{-n/6}$$
I could check the case  $n=1,2$ by computation using wolfram.
I don't have any concrete proof.
 A: Pick a number $a \in (1, \pi)$ whose value is to be determined later. Then
$$ \int_{a}^{\infty} \left| \frac{\sin x}{x} \right|^n \, \mathrm{d}x
\leq \int_{a}^{\infty} \frac{\mathrm{d}x}{x^n}
= \frac{a^{1-n}}{n-1} \tag{1} $$
Next, for $0 < x < \pi$, the Weierstrass factorization for $\sin x$ gives
$$ 0 \leq \frac{\sin x}{x}
= \prod_{k=1}^{\infty} \left( 1 - \frac{x^2}{k^2\pi^2}\right)
\leq \prod_{k=1}^{\infty} e^{-x^2/(k\pi)^2}
= e^{-x^2/6}, $$
and so, it follows that
$$ \int_{1}^{a} \left( \frac{\sin x}{x} \right)^n \, \mathrm{d}x
\leq \int_{1}^{a} e^{-nx^2/6} \, \mathrm{d}x
\leq (a - 1)e^{-n/6}. \tag{2} $$
The sum of the bounds $\text{(1)}$ and $\text{(2)}$ turns to be minimized at $a = e^{1/6}$, and we use this choice hereafter. Then combining $\text{(1)}$ and $\text{(2)}$ together,
$$ \int_{1}^{\infty} \left( \frac{\sin x}{x} \right)^n \, \mathrm{d}x
\leq e^{-n/6} \left( \frac{n}{n-1}e^{1/6} - 1 \right) $$
We can check that $\frac{3}{2}e^{1/6} - 1 < 1$, and so, this confirms that the inequality in the question holds for $n \geq 3$. The cases $n = 1, 2$ can be checked individually.
