# showing $f(x) =\frac{(\ln(x))^{\alpha}}{(x-1)^{\beta}}$ is integrable for certain $\alpha > 0$ and $\beta \in \mathbb{R}$

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

$$f$$ 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 $$\int_0^1\frac{1}{x^\gamma}dx<\infty \Leftrightarrow \gamma<1,\quad\quad \int_1^\infty\frac{1}{x^\gamma}dx<\infty \Leftrightarrow \gamma>1.$$

• 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 $x \to 1$. Nov 21, 2021 at 22:14
• $f(x)=\mathcal O((x-1)^{-\beta}$ 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 $\gamma>0$. Nov 22, 2021 at 8:11
• okay and how would I go about doing that? Nov 22, 2021 at 8:56
• I mean $(\ln x)^{\alpha} / (x-1)^{\beta - \beta'} \leq 1$, I'm not able to edit my comment Nov 23, 2021 at 11:21
• Yes. It then gives you $f(x)\le C(x-1)^{-\beta'}$. Nov 23, 2021 at 18:25