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Suppose that $p\geq 1.$ This post shows that the following integral $$\int_0^\infty \frac{1}{x ((\ln x)^2+1)^p} dx$$ converges for $p.$

Question: Suppose that $q\geq 1$ such that $q \neq p.$ Is it true that $$\int_0^\infty \left|\frac{1}{x^{1/p} ((\ln x)^2+1)}\right|^q dx$$ is divergent?

I am aiming to obtain a function $g$ such that $$g(x) \leq \frac{1}{x^{q/p}((\ln x)^2+1)^q}$$ and $\int_0^\infty g(x)\,dx=\infty.$ However, I fail to obtain such $g.$

Any hint would be appreciated.

Masacroso
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Idonknow
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  • try to write the integral as the sum of two improper integrals, one in $[0,1]$ and the other in $[1,\infty)$, and search such $g$ (or an upper bound) for each case. – Masacroso Aug 23 '17 at 07:21
  • The substitution $x = e^u$ gives $$ \int_{0}^{\infty} \left( \frac{1}{x^{1/p}(1 + \log^2 x)} \right)^q , dx = \int_{-\infty}^{\infty} \frac{e^{\alpha u}}{(1+u^2)^q} , du$$ where $\alpha = 1 - \frac{q}{p}$. Certainly this diverges for all $\alpha \neq 0$. – Sangchul Lee Aug 23 '17 at 09:12
  • How to show your last integral diverges? Actually I also got to your last integral. However, I could not proceed. Any hint? – Idonknow Aug 23 '17 at 09:15
  • @SangchulLee: If I use the fact that $e^{\alpha u} \geq \alpha u,$ then I can show that the integral diverges for $q \neq 1.$ But I have no way to eliminate the case $q = 1.$ – Idonknow Aug 23 '17 at 09:27
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    Exponential beats any polynomial growth. If you need an explicit lower bound, notice that $e^x \geq \frac{x^n}{n!}$ for all $x \geq 0$ and $n \geq 0$. – Sangchul Lee Aug 23 '17 at 09:35

1 Answers1

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Partial answer:

Taking $\ln x=u$, $$I=\int_0^\infty \left|\frac{1}{x^{1/p}(1+(\ln x)^2)}\right|^q dx=\int_{-\infty}^\infty\frac{e^{(1-q/p)u}}{(1+u^2)^q}du=\int_{-\pi/2}^{\pi/2}e^{r\tan\theta} \cos^{2q-2}\theta d\theta\\=2\int_{0}^{\pi/2}\cosh(r\tan \theta)\cos^{2q-2}\theta d\theta $$ where $r=1-p/q$. One way to proceed now is to expand the $\cosh$ term in series and then use the $\Gamma$ function formulas.

Samrat Mukhopadhyay
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