For what value of $p>0$, the function $f(x)=\sin(x)/x$ lies in $L^p(0,\infty)$.
I think this function is in $L^p$ for $0<p<1$. How to check for $p\geq1$?
Can anyone help me?
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I think for $p>1$ this is easy to see by the integral test. For $p > 1$ and $x$ away from $0$, $\left|\frac{\sin x}{x}\right|^p \le \frac{1}{|x|^p}$. The latter is integrable, so your function will be integrable. The other case ($0 < p \le 1$) is not quite as simple. You can get an idea from here: http://math.stackexchange.com/questions/225439/showing-frac-sin-xx-is-not-lebesgue-integrable-on-mathbbr-ge-0 – Cameron Williams Feb 04 '15 at 04:40
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To be precise, the function is not in $L^p$ for $p \leq 1$, but it is in $L^p$ for $p>1$. The point is that the singularity at $0$ is removable (because of $\sin(x)/x \to 1$ for $t \to 0$), so the interesting thing is the behaviour at $\infty$. Here, $f(x)$ decays at least as fast as $x^{-1}$, which allows to show $f \in L^p$ for $p>1$. For the case $p\leq 1$, the link provided by @CameronWilliams should help you. – PhoemueX Feb 04 '15 at 09:20
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Can you show both case me details – user145993 Feb 04 '15 at 12:21