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I need to determine the values of p for which this integral converges conditionally and absolutely.

$$\int_{0}^{\infty} \dfrac{\sin(x)}{x^p} dx $$

I think the interval for conditional convergence is $0 < x < 2$ and for absolute convergence the interval is probably $0 < x < 1$. I'm guessing that I need to somehow compare it with the $\dfrac{1}{x^p}$ integral, but I am not sure exactly where to begin and how to logically proceed. I first thought about dividing the integral into two separate improper integrals, with one of them integrating from $0$ to $1$ and the other from $1$ to infinity, but I do not know how to continue from that point on.

egreg
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  • This is similar to: http://math.stackexchange.com/questions/358434/to-determine-whether-the-integral-int-0-infty-frac-sinaxbxp-m – Baby Dragon Oct 06 '13 at 01:24
  • Only difference is that in this question absolute convergence is also addressed. –  Oct 06 '13 at 01:52
  • It is really strange that this question is closed as a duplicate of the other one asked and answered 6 months later. This contradicts to the statement "This question has been asked before and already has an answer." – user Mar 19 '19 at 19:45
  • @user: This was discussed on meta before. The point of a duplicate is not to establish priority, but rather to comment on generality of the question and/or the answers. If you still disagree, you can post on the meta thread for requests for reopen votes why you think this should go the other way around (or perhaps as a separate thread, since the target here is a duplicate of a third question, and I'm sure there might be more). – Asaf Karagila Mar 26 '19 at 07:32
  • @AsafKaragila After the other thread was closed (on my request) as duplicate of a third question the particular problem has lost actuality, though the word "before" in the explanation of closing duplicates cannot be in my opinion interpreted in the way you suggest (if one considers time as an ordered set). – user Mar 26 '19 at 12:41
  • @user: Time is not linear when duplicates are concerned. It is not a flat circle either, despite what Matthew McConaughey might have told you. – Asaf Karagila Mar 26 '19 at 12:54
  • @AsafKaragila "Time is not linear"... I will try to memorize this masterpiece. Thank you. – user Mar 26 '19 at 13:00
  • @user: If you are going to take my words out of context I am sure you can do better. For example, you could take the last part of the previous sentence and conclude that I not being sarcastic. – Asaf Karagila Mar 26 '19 at 13:03

2 Answers2

2

(1) For $\,p=0\,$ the integral diverges.

(2) Lets prove that the integral converges for $\,p\in(0,2)$. $$\left|\frac{\sin{x}}{x^p}\right|\leqslant \frac{1}{x^{p-1}}\quad ,\,x>0 \quad \Rightarrow\quad \int_{0}^{\lambda}{\left|\frac{\sin{x}}{x^p}\right|dx}\leqslant \int_{0}^{\lambda}{\frac{dx}{x^{p-1}}}$$ which converges for $\enspace p-1< 1\enspace \Leftrightarrow \enspace p<2$, which is true. Dirichlet's test implies that $\int_{\lambda}^{\infty}{\frac{\sin{x}}{x^p}dx}$ converges for $\,p>0$.

(3) Lets prove that the integral diverges for $\,p\geqslant2$.

$$\lim_{x \rightarrow 0}\frac{x^{p-1}}{\sin{x}}=0\,(or\,1,\,if\,p=2)\quad\Rightarrow\quad \frac{\sin{x}}{x^p}\geqslant \frac{17}{x}\quad ,x\in(0,\varepsilon)$$ $$\Rightarrow\quad\int_{0}^{\varepsilon}{\frac{\sin{x}}{x^p}dx}\geqslant 17\int_{0}^{\varepsilon}{\frac{dx}{x}}$$ which diverges.

Therefore, the whole integral diverges for $\,p\geqslant2\enspace$because $\int_{0}^{\infty}=\int_{0}^{\varepsilon}+\int_{\varepsilon}^{\infty}$.

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user145542
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1

Hints: The following facts are useful for your problem. Close to zero the integrand behaves as

$$ \frac{\sin x}{x^p} \sim \frac{x}{x^p}, $$

and

$$ \Bigg| \frac{\sin(x)}{x^p}\Bigg|\leq \frac{1}{x^p}. $$

Mhenni Benghorbal
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