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$$ \int_{0^+}^{1}{\frac{x^a\ln(x^b)}{e^x-1}}dx,\quad \forall a,b \in \mathbb{R},a\ge 1 $$

It's from a final exam. I just can't find a proper function to prove convergence or divergence.

For $b=0 \rightarrow \frac{x^a\ln(x^b)}{e^x-1}=0 \therefore$ the integral converges.

I know that $ \int_{0^+}^{1}{x^a\ln(x^b)}dx$ converges (although not useful for LCT).

Assuming that $\int_{0^+}^{1}{\frac{\ln(x)}{e^x-1}}dx$ converges:

For $b\lt0 \rightarrow \ln(x^b)\gt0\therefore$ I can use LCT. Is correct to say that if $b\gt0$ I can multiply everything by $-1$ and say that convergence is also valid in this case? Then I can take $b$ out of the integral and check with LCT:

$$\lim_{x\rightarrow0^+}{\frac{\frac{x^a\ln(x)}{e^x-1}}{\frac{\ln(x)}{e^x-1}}}=\lim_{x\rightarrow0^+}{x^a}=0$$

Yet again I say that if $\int_{0^+}^{1}{\frac{\ln(x)}{e^x-1}}dx$ converges then $-\int_{0^+}^{1}{\frac{\ln(1/x)}{e^x-1}}dx$ also converges. Now I can use LCT again, but got stuck proving this. I thought about using DCT for $\int_{0^+}^{1}{\frac{\ln(1/x)}{e^x-1}}dx\le\int_{0^+}^{1}{\frac{1/x}{e^x-1}}dx\le\int_{0^+}^{1}{\frac{1}{e^x-1}}dx$, but again couldn't prove convergence on anyone.

Can you guide me on which way to take?

Gary
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Pedro Giuttari
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  • Just to be clear, this is from an old final exam that you're using to study, right? – Robert Shore Jan 25 '22 at 01:31
  • Hahaha, of course. My real final is in a month. This one is 5 years old. – Pedro Giuttari Jan 25 '22 at 01:48
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    Why don't you use $\ln(x^b)=b\ln x$ to get rid of the $b$? It does not play any role in the problem. – Gary Jan 25 '22 at 01:49
  • I did took out b (when I said "I can take b out of the integral"). But for using LCT the integrand must be always positive, right?. That's where b plays a role (I think). – Pedro Giuttari Jan 25 '22 at 01:51
  • I mean do it at the very beginning and do not talk about it at all. Btw, "proof" is the noun and "prove" is the verb. I fixed it. – Gary Jan 25 '22 at 01:51
  • Thanks, english is my second language. – Pedro Giuttari Jan 25 '22 at 01:53
  • Obviously the convergence problem is at $x=0$, and $$ \frac{{x^a \ln (x^b )}}{{e^x - 1}} = b\frac{x}{{e^x - 1}}x^{a - 1} \ln x \sim bx^{a - 1} \ln x $$ as $x\to 0+$. This will show you that the requirement is $a>0$. – Gary Jan 25 '22 at 01:54
  • Thanks. I'll see what comes out of that and get back to you. – Pedro Giuttari Jan 25 '22 at 01:57
  • Oh, it is given that $a\geq 1$. Then it is even easier: $$ \left| {\frac{{x^a \ln (x^b )}}{{e^x - 1}}} \right| = \left| b \right|\frac{x}{{e^x - 1}}x^{a - 1} \left| {\ln x} \right| \le \left| b \right|x^{a - 1} \left| {\ln x} \right| \le \left| b \right|\left| {\ln x} \right| $$ for $0<x<1$. – Gary Jan 25 '22 at 02:24
  • This works!. Thanks Gary. Please post it, so I can upvote it as the correct answer. – Pedro Giuttari Jan 25 '22 at 03:38

1 Answers1

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If $a\geq 1$ and $b\in \mathbb R$, then \begin{align*} \left| {\int_0^1 {\frac{{x^a \ln (x^b )}}{{e^x - 1}}dx} } \right| &= \int_0^1 { \frac{{x^a \left| {\ln (x^b )} \right|}}{{e^x - 1}}dx} = \left| b \right|\int_0^1 {\frac{x}{{e^x - 1}}x^{a - 1} \left| {\ln x} \right|dx} \\ & \le \left| b \right|\int_0^1 {\left| {\ln x} \right|dx} = \left| b \right|. \end{align*}

Gary
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