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Does the integral

$$\int\limits_0^\infty \frac{\sin^{2}x}{x^{2}\ln(1+\sqrt x)} dx$$

converge?

It's easy to check that $\int\limits_1^{\infty}\frac{\sin^{2}x}{x^{2}\ln(1+\sqrt x)} dx$ does converge, but I couldn't find the right method for either proving or disproving that $\int\limits_0^1 \frac{\sin^{2}x}{x^{2}\ln(1+\sqrt x)} dx$ converges.

David G. Stork
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gbi1977
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1 Answers1

2

We might check some equivalent for $x\mapsto \frac{\sin^2(x)}{x^2 \ln(1+\sqrt{x})}$ near zero.

hint: $$ \ln(1+x) = x + o(x)$$

A. PI
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  • the first thing I tried is: since $\ln (1+\sqrt x)<\sqrt x$, then $\frac{1}{\sqrt x}<\frac{1}{\ln (1+\sqrt x)}$, but it's not valid the use the comparsion test like that. – gbi1977 Jul 27 '17 at 18:17
  • @gbi1977 $$\frac{\sqrt x}{1+\sqrt x}\le \log(1+\sqrt x)\le \sqrt x$$ – Mark Viola Jul 27 '17 at 18:19
  • @gbi1977 the inequality is true (out of zero) but is not useful to prove the convergence – A. PI Jul 27 '17 at 18:23
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    $\frac{1}{log(1+\sqrt{x})} \leq \frac{2}{\sqrt{x}}$ on $]0,1]$ – A. PI Jul 27 '17 at 18:25
  • @MarkViola I was not familiar with the left side inequality... thanks – gbi1977 Jul 27 '17 at 18:26
  • @A.MONNET where did you derive that? – gbi1977 Jul 27 '17 at 18:28
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    @gbi1977 See This Answer, which develops the left side estimate. And for $x\in (0,1)$, we see that since $1+\sqrt x\le 2$, $\frac{1}{\log(1+\sqrt x)}\le \frac{1+\sqrt x}{\sqrt x}\le \frac2{\sqrt x}$ as A. Monnet commented. – Mark Viola Jul 27 '17 at 18:28
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    If you don't like the 2, just put any constant $c>0$ and study the function $f(x) := c\log(1+\sqrt{x}) - \sqrt{x}$ on $]0,1]$ then choose $c$ to get the good inequality on $]0,1]$ – A. PI Jul 27 '17 at 18:36