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For what values of $\alpha > 0$ and $\beta \in \mathbb{R}$ is the function \begin{equation} f:(1, +\infty) \rightarrow \mathbb{R}:f(x) =\frac{(\ln(x))^{\alpha}}{(x-1)^{\beta}} \end{equation} integrable?

I think I have to show that $f(x) = \Omega(h(x))$ for some function $h(x)$ in the limit $x \rightarrow 1$ and $x \rightarrow + \infty$.

Since $\lim_{x \to 1}\frac{\ln x}{x} = 1$, we have that $\lim_{x \to 1}\left(\frac{\ln x}{x}\right)^{\beta}$ = 1 for $\beta \geq 0$. I think we should use this, but I don't know how to go further about proving this...

Could anyone help?

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$f$ is integrable if and only if $\beta-\alpha<1$ and $\beta>1$.

For if part, split integral into two parts:

$$\int_{1}^\infty f(x)dx=\underbrace{\int_{1}^2f(x)dx}_{=I_1} +\underbrace{\int_{2}^\infty f(x)dx}_{=I_2}.$$

On $[1,2]$, $f(x)\le C(x-1)^{\alpha-\beta}$ using the limit that you have. Then $I_1$ is finite since $\beta-\alpha<1$. On $[2,\infty)$, there exists $1<\beta'<\beta$ such that $f(x)\le C (x-1)^{-\beta'}$ so $I_2$ is finite.

For the only if part, suppose either

  1. $\beta-\alpha\ge1$. Then $f(x)>C(x-1)^{-1}$ on $[1,2]$ so $I_1$ diverges.
  2. $\beta\ge1$. Then $f(x)>C(x-1)^{-1}$ on $[2,\infty)$ so the $I_2$ diverges.

The main idea is to control the integral near $1$ and near $\infty$. We have only used that $$\int_0^1\frac{1}{x^\gamma}dx<\infty \Leftrightarrow \gamma<1,\quad\quad \int_1^\infty\frac{1}{x^\gamma}dx<\infty \Leftrightarrow \gamma>1.$$

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  • $\begingroup$ can we somehow prove that $f(x) = \Theta\left(\frac{1}{(x-1)^{\beta}}\right)$ for $x \to \infty$ ? If this is true, I think I can prove it since I was able to prove that $f(x) = \Theta\left(\frac{1}{(x-1)^{\beta-\alpha}}\right)$ for $x \to 1$. $\endgroup$ Nov 21, 2021 at 22:14
  • $\begingroup$ $f(x)=\mathcal O((x-1)^{-\beta}$ is not true because of the $\log x$ factor. Instead, you can show that for any $\beta'<\beta$, $f(x)=\mathcal O((x-1)^{-\beta'}$ since $\log x$ is dominated by any $x^\gamma$ for $\gamma>0$. $\endgroup$
    – kcjkgs
    Nov 22, 2021 at 8:11
  • $\begingroup$ okay and how would I go about doing that? $\endgroup$ Nov 22, 2021 at 8:56
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    $\begingroup$ I mean $(\ln x)^{\alpha} / (x-1)^{\beta - \beta'} \leq 1$, I'm not able to edit my comment $\endgroup$ Nov 23, 2021 at 11:21
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    $\begingroup$ Yes. It then gives you $f(x)\le C(x-1)^{-\beta'}$. $\endgroup$
    – kcjkgs
    Nov 23, 2021 at 18:25

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