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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.

Guy Fsone
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  • Maybe helpful link – S.H.W Nov 02 '20 at 22:51
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    It can't be true. We know that $$\int_{0}^{\infty}\frac{\sin x}{x},\mathrm{d}x=\int_{0}^{\infty}\biggl(\frac{\sin x}{x}\biggr)^2,\mathrm{d}x=\frac{\pi}{2}.$$ – Sangchul Lee Nov 02 '20 at 22:54
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    Moreover, the Laplace's method gives $$\int_{0}^{\infty}\biggl(\frac{\sin x}{x}\biggr)^n,\mathrm{d}x\sim\frac{1}{\sqrt{n}}\int_{0}^{\infty}e^{-x^2/6},\mathrm{d}x=\sqrt{\frac{3\pi}{2n}}.$$ It cannot decay exponentially fast. – Sangchul Lee Nov 02 '20 at 23:00
  • Sorry I have corrected – Guy Fsone Nov 02 '20 at 23:00
  • There is an exact result for the integral of $\operatorname{sinc}^n$ over $\mathbb{R}_+$. https://math.stackexchange.com/questions/307510/a-sine-integral-int-0-infty-left-frac-sin-x-x-rightn-mathrmdx/307837#307837 – K.defaoite Nov 02 '20 at 23:08
  • @K.defaoite check again – Guy Fsone Nov 02 '20 at 23:19

1 Answers1

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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.

Sangchul Lee
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