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Problem: does the integral $\displaystyle\int_0^1 \frac{\sin\big(\frac1{x^2}\big)}{\frac\pi2 -\ \rm{arccotg}\ x}$ converge?

I know that the function is continuous on $(0,1]$, that the problem is at $x = 0$ and that I'll have to use Abel's (or Dirichlet's) convergence criteria. But I cannot find the functions $g, h$ so that $f = gh$, $g$ is monotonic and $f$ is "known" (in the sense that antiderivative is known or $\int_0^1 f$ converges.)

Could anyone provide a hint?

Edit: One of my previous attempts was to write

$$ f(x) = (-2)\frac{\sin{\frac1{x^2}}}{x^3}\frac{x^3}{-2\big(\frac\pi2 - {\rm arccotg}\ x\big)} $$

Then I can prove that the first factor converges (using $t = \frac1{x^2}$). The problem is that I do not know how the prove the monotonicity of the second factor.

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  • $\begingroup$ HINT: Split your integral in both and treat the integrand on a neighborhood of $0$ by Taylor expansion. $\endgroup$ – Nicolas Sep 13 '15 at 8:09
  • $\begingroup$ Thanks! But how will that work for me? The function changes signs at $(0, 1)$, therefore, I cannot use the limit test. And the expansion can only be done for ${\rm arccotg} x$. $\endgroup$ – David Sep 13 '15 at 8:12
  • $\begingroup$ I think there is a typo: shouldn't it be "or $\int_0^1 f$ converges"? $\endgroup$ – rubik Sep 13 '15 at 8:18
  • $\begingroup$ Oh, sure. I'm sorry, I'll fix that. $\endgroup$ – David Sep 13 '15 at 8:20
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    $\begingroup$ @eudes, that may be it! That will simplify the calculation of the derivative. Great! $\endgroup$ – David Sep 13 '15 at 8:33
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First of all, note that $\frac \pi2 - \operatorname{arc ctg} x = \operatorname{arc tg} x$ for $x\geq 0$. If anything, this makes the problem shorter to write, and maybe easier to think of. Also, bear in mind that $\operatorname{arc tg} x\approx x$ near $0$, more precisely: $$ \lim_{x\to 0 } \frac{\operatorname{arc tg} x}x = 1 \quad \text{and} \quad \frac 12 x \stackrel{(\star)}< \operatorname{arc tg} x < x \quad \text{for } x\in(0, \varepsilon) $$ (in fact at least for $x\in(0, 2\pi/3)$ and actually for $x\in (0,0.7421\dots\pi)$; if you don't know these, see the remark at the end).

Now, if we write the integrand as you suggest (and using my little note, and simplifying a bit): $$ \frac{\sin{\frac1{x^2}}}{x^3}\frac{x^3}{\operatorname{arc tg} x} $$ and you're asking about the monotonicity of the second factor, then we should try the derivative. I'll leave the calculation as an exercise – let's examine the result: $$ \big(\frac{x^3}{\operatorname{arc tg} x}\big)' = \frac{x^2}{(\operatorname{arc tg}x)^2} \bigg(3 \operatorname{arc tg} x - \frac x{x^2+1}\bigg). $$ Because of the inequalities: $(\star)$ and $x^2 + 1 > 1$: $$ 3 \operatorname{arc tg} x - \frac x{x^2+1} > \frac 32 x - x = \frac 12 x > 0 $$ for $x\in(0,\varepsilon)$, so the second factor is increasing in $(0,\varepsilon)$ (in fact, in $(0, \infty)$), which should yield the convergence of the integral $\displaystyle\int_0^\varepsilon (\dots)$.

Remark. The limit follows from the limit $\frac{\operatorname{tg} x}x\to 1$ and further from $\frac{\sin x}x\to 1$, which is classic, I think. The inequality $\operatorname{arc tg}x < x$ follows from $\operatorname{tg}x > x$, which I suppose is also classic. Finally, given the limit, it must be $\operatorname{arc tg}x > \frac 12 x$ close to zero: for some $x\in(0,\varepsilon)$. The coefficient $\frac 12$ isn't special: anything in $(\frac 13, 1)$ will do fine here, but $\frac 12$ came first to my mind.

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  • $\begingroup$ @David Can't believe it took me well over hour to write this. You in turn are very quick. Thanks for accepting. I fact, what I actually did were only two little steps, I just expanded it in the answer, so that it is complete and helpful for the "general public". You already did most of the work. Happy to help anyway! $\endgroup$ – eudes Sep 13 '15 at 11:41
  • $\begingroup$ never mind about that, what I regard the best is that I had somenone to bear the burden with me for a while :). You have a nice time! $\endgroup$ – David Sep 13 '15 at 12:11
  • $\begingroup$ @David Well, this pleasure might continue. I see a problem, don't know yet how serious it is. I couldn't understand what you wrote: that you can prove convergence for the 1st factor. In fact, WolframAlpha would claim it doesn't converge. So, are you sure about that? $\endgroup$ – eudes Sep 13 '15 at 12:16
  • $\begingroup$ @David Ok, so it turns I don't remember the Abel-Dirichlet criterion :/ It seems fine. Although still I wouldn't say that the 1st factor "converges". $\endgroup$ – eudes Sep 13 '15 at 12:23
  • $\begingroup$ @David Sorry for disturbing and have a nice Sunday! $\endgroup$ – eudes Sep 13 '15 at 12:35

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