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I can't prove that for $\alpha>0$, $I_\alpha=\int_{0}^{+\infty}\frac{t-\sin{t}}{t^\alpha}\,dt$ converges iff $\alpha\in (2,4)$. Here's my attempt:

1) An easy remark but important: as $\sin{t}\le t$, we're dealing with a positive function.

2) I splited the integral to $\int_{0}^{1}\frac{t-\sin{t}}{t^\alpha}\,dt$ and $\int_{1}^{+\infty}\frac{t-\sin{t}}{t^\alpha}\,dt$ and use the inequality $\frac{t^3}{3!}-\frac{t^5}{5!}\le t-\sin{t}\le \frac{t^3}{3!}$ but it didn't work.

3) This is the only thing that lead me to a part of the answer: I noticed that $\int_0^{+\infty}\frac{dt}{t^{\alpha-1}}$ is always divergent. Thus, if $I_\alpha$ converges, $\int_0^\infty\frac{\sin{t}}{t^\alpha}\,dt$ diverges. But we know that $\int_1^\infty\frac{sin{t}}{t^\alpha}\,dt$ converges for any $\alpha>0$. Thus $\int_0^\infty\frac{\sin{t}}{t^\alpha}\,dt$ diverges iff $\int_0^1\frac{\sin{t}}{t^\alpha}\,dt$ diverges. Since $\sin{t}\sim t$ near $0^+$, both positive, $\int_0^1\frac{\sin{t}}{t^\alpha}\,dt$ diverges iff $\alpha>2$. Hence we showed that $(I_\alpha\,\text{converges})\Rightarrow\alpha>2$

Could you please help me? Thank you in advance!

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2 Answers 2

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The idea to split the integral is fine.

First, note that the integral $I_1$ as given by

$$I_1=\int_0^1 \frac{t-\sin(t)}{t^\alpha}\,dt$$

converges for $\alpha<4$ (and diverges for $\alpha \ge 4)$ since the integrand is $O\left(t^{3-\alpha}\right)$ as $t \to 0$.

Next, note that the integral $I_2$ as given by

$$\int_1^\infty \frac{t-\sin(t)}{t^\alpha}\,dt$$

converges for $\alpha >2$ (and diverges for $\alpha \le 2$) since the integrand is $O\left(t^{1-\alpha}\right)$ as $t\to \infty$.

Putting it together, we have that the integral of interest $I=I_1+I_2$ converges for all $\alpha \in (2,4)$ and diverges elsewhere.

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  • $\begingroup$ Firstly thank you very much for your answer. (I didn't write it right after voting because I had some errands to do). Secondly, I'm so sorry for that mistake and I apologize for it! Indeed, when I saw both of your answers, I did the "answered xxx ago" above your names. It was written "17h" for your answer and "16h" for the other answer. I just don't know why I took the order of posting the other way hahaha! My bad, making such a ridiculous mistake! Also my bad I didn't give much attention to the $O$ there. Thank you again! $\endgroup$ Jul 15, 2016 at 15:38
  • $\begingroup$ You're quite welcome. My pleasure. And thank you for the best vote! Very much appreciated. -Mark $\endgroup$
    – Mark Viola
    Jul 15, 2016 at 15:45
  • $\begingroup$ You deserve it! :) $\endgroup$ Jul 15, 2016 at 15:55
  • $\begingroup$ Let me know if there are other math questions you have and I'll do my best to help you. -Mark $\endgroup$
    – Mark Viola
    Jul 15, 2016 at 17:14
  • $\begingroup$ That's so kind. Thank you very much! $\endgroup$ Jul 16, 2016 at 14:48
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Split the integral $$\int_{0}^{+\infty}f_\alpha(t)\,dt = \int_{0}^{1}f_\alpha(t)\,dt + \int_{1}^{+\infty}f_\alpha(t)\,dt$$ where $f_\alpha(t) = \frac{t-\sin{t}}{t^\alpha}$.

For small $t$ (i.e. $t \in [0, 1]$), we have that $$f_\alpha(t) =-\frac{\frac{t^3}{6} + O(t^5)}{t^\alpha} \sim -t^{3-\alpha}.$$

Then:

$$-\int_{0}^{1} t^{3-\alpha}\,dt = -\int_{0}^{1} \frac{1}{t^{\alpha-3}}\,dt$$ is convergent if $$\alpha-3 < 1 \Rightarrow \alpha < 4.$$

For large $t$ (i.e. $t > 1$), we have that $$f_\alpha(t) \sim \frac{t}{t^{\alpha}} = t^{1-\alpha}.$$

In this case:

$$\int_{1}^{+\infty} t^{1-\alpha}dt = \int_{1}^{+\infty} \frac{1}{t^{\alpha-1}}dt$$ converges when

$$\alpha-1 > 1 \Rightarrow \alpha > 2.$$

Joining the two condition, you get that the integral converges when $$\alpha \in (2, 4).$$

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  • $\begingroup$ Thank you very much for your answer! $\endgroup$ Jul 15, 2016 at 15:40

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