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I want to find all pairs $(a,b)$ where $a,b\in \mathbb{R}$ such that $\int_{1}^{\infty}\frac{(\ln x)^{b}}{x^{a}}dx$ is finite.

I found some parts of the solution. For instance, when $b<-2$ and $a>1$ or when $b=1$ and $a>2$, this integral is finite. But I don't know how to analyze it in general. Is there any way to see all $(a,b) $ easier?

MPW
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Tuğba Yesin
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  • May be this helps https://math.stackexchange.com/questions/3702921/for-which-values-of-p-q-does-the-integral-%e2%88%ab10xpln1xqdx-converge/3702993#3702993 – zkutch Jun 30 '20 at 14:30
  • A related post here https://math.stackexchange.com/questions/3723448/find-the-real-values-of-p-and-q-such-that-following-integral-converges/3723795#3723795 – Mittens Jun 30 '20 at 14:34

2 Answers2

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The convergence of this integral is related to convergence of Bertrand's Series through application the integral test. But we need to be careful due to the possible singularity at $x=1^+$ arising from $\log^b(x)$ for $b<0$.

One way to see things a bit clearer is to enforce the substitution $x\mapsto e^x$. Proceeding, we obtain

$$\int_1^\infty \frac{\log^b(x)}{x^a}\,dx=\int_0^\infty x^be^{(1-a)x}\,dx\tag 1$$

And now the singularity at $x=1$ for $b<0$ of the integrand on the left-hand side of $(1)$ is more transparent by looking at the singularity at $x=0$ for $b<0$ of the integrand on the right-hand side of $(1)$.

From the right-hand side of $(1)$, it is straightforward to see that the integral diverges for $a\le 1$ and converges for $a>1$ and $b>-1$.

Mark Viola
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  • I am trying to understand your explanation but I couldn't see the result for just the case $a>1$ and $b<0$. Thank you for your interest. – Tuğba Yesin Jul 02 '20 at 08:26
  • For $a>1$, the exponential decay overwhelms any possible power of $x$. So, the integral $\int_1^\infty x^be^{-x},dx$ converges. If $-1<b<0$, then for small $x$ the integrand behaves like $\frac1{x^{|b|}}$ and $\int_0^1 \frac1{x^{|b|}},dx=\left.\left(\frac{x^{1-|b|}}{1-|b|}\right)\right|_0^1=\frac1{1-|b|}$ exists. – Mark Viola Jul 02 '20 at 14:30
  • I understood all cases. Thank you. – Tuğba Yesin Jul 04 '20 at 10:13
  • You're welcome. My pleasure. And pleased to hear that you have it now. – Mark Viola Jul 04 '20 at 13:57
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hint

Let $$f(x)=\frac{(\ln(x))^b}{x^a}$$ Assume $ b\ge 0$. $ f $ is continuous at $[1,+\infty)$, so it is locally integrable and positive at $ [1,+\infty)$.

If $ a>1$ then catch $\alpha \in (1,a)$.

$$\lim_{x\to+\infty}x^{\alpha}f(x)=0$$ $$\implies 0\le f(x) \le \frac{1}{x^\alpha} $$ for $ x$ great enough. thus the integral converges.

If $ a<1 $, take $ \beta\in (a,1)$

$$\lim_{x\to+\infty}x^\beta f(x)=+\infty$$ $$\implies \frac{1}{x^\beta}\le f(x)$$ for $ x $ great enough. then the integral diverges.

If $ a=1$, then for $ X>1$

$$\int_1^Xf(x)dx=\Bigl[\frac{(\ln(x))^{b+1}}{b+1}\Bigr]_1^X$$

the integral converges $ \iff b+1<0$

I let for you the cases $ a=1$ and $ b<0$.

hamam_Abdallah
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  • How on earth does the integral converge for $a=1$ and any value of $b$? If $a=1$, $\lim_{x\to\infty} \log^{b+1}(x)=\infty$ for $b+1>0$, and $\lim_{x\to 1^+}\log^{b+1}(x)=\infty$ for $b+1<0$. For $a=1$ and $b=1$, $\lim_{x\to\infty}\log^2(x)=\infty$. So, the integral fails to converge for $a\le 1$ for all $b$. – Mark Viola Jun 30 '20 at 15:58
  • Your post has fatal flaws and should be either corrected or deleted as it is misleading. – Mark Viola Jul 01 '20 at 19:28