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Let $n \in N$.

How to compare two integrals: $$ I_1=\int_0^{\infty}\left(\frac{\sin t}{t}\right)^n dt \quad \text{and} \quad I_2=\int_0^{\pi}\left(\frac{\sin t}{t}\right)^n dt\,\, ? $$

I've beet trying to compare them for some particular $n$. It seems, that $I_1>I_2$ for $n-\text{even}$ and $I_2>I_1$ for odd $n$.

Help me please to prove this statement for any $n\in N$.

Thank you.

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  • $\begingroup$ For $I_1$, try to split the interval as $[0,\infty)=\cup_{n=0}^{\infty} [n\pi, (n+1)\pi) $. $\endgroup$ – Mhenni Benghorbal Feb 5 '14 at 21:18
  • $\begingroup$ For even $n$ we have $f_n\geq 0$. Hence $I_1\geq I_2$ in this case, actually $I_1=+\infty$ in this case. Here $$f_n(x) = \left(\frac{\sin x}{x}\right)^n$$ $\endgroup$ – AD. Feb 6 '14 at 8:18

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