What values of alpha make the integral converge?

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Homework Help Overview

The discussion revolves around determining the values of alpha for which the integral \(\int_0^1 x^\alpha \ln(x) \, dx\) converges. Participants are exploring the conditions under which this integral behaves properly as \(x\) approaches 0.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to identify appropriate tests for convergence and are questioning how to calculate limits as \(x\) approaches 0. Some express uncertainty about the integration process and the implications of the limits involved.

Discussion Status

The discussion is ongoing, with various participants sharing insights and hints regarding the limits and conditions for convergence. There is an acknowledgment of the need for clarification on how to derive specific values of alpha, but no consensus has been reached yet.

Contextual Notes

Some participants mention the use of L'Hospital's rule and the importance of continuity in the context of evaluating the integral, indicating that there may be constraints or assumptions that need to be addressed further.

gipc
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Hello,
I need to find the values of alpha for which the following integral converges:
[tex]\int x^\alpha*ln(x)[/tex] the integral is from 0 to 1.

I don't really know which test should I use or how to calculate the limit of the integral as x->o+
 
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Please see attachment for some insight.
 

Attachments

  • integral.JPG
    integral.JPG
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I still can't see how to find the alphas given that statement after you integrated
 
This might not be the correct approach to the problem, I was just trying to give you some insight that perhaps would be helpful. Hopefully someone will clarify this further.
 
I'm sorry for the bump. Can anyone please help?
 
It's really all in proceeding from the hint given in the attachment.

You end up with

[tex]\left(\frac{1}{\alpha+1}x^{\alpha+1}\ln(x)\right)\bigg|_0^1 - \left(\frac{1}{(\alpha+1)^2}x^{\alpha+1}\right)\bigg|_0^1[/tex]

This will exist if (and only if) the two limits

[tex]\lim_{x\to 0^+}\frac{1}{\alpha+1}x^{\alpha+1}\ln(x),\quad <br /> \lim_{x\to 0^+}\frac{1}{(\alpha+1)^2}x^{\alpha+1}[/tex]

both exist. Use L'Hospital's rule for the first (write it with [tex]x^{-\alpha-1}[/tex]) on bottom, and the second is immediate. You should get the same answer for both
limits working out.
 
gipc said:
Hello,
I need to find the values of alpha for which the following integral converges:
[tex]\int x^\alpha*ln(x)[/tex] the integral is from 0 to 1.

I don't really know which test should I use or how to calculate the limit of the integral as x->o+

If you have general integral

[tex]\int_{a}^{b} f(x) dx[/tex] and want to find the limit then here is where you do

Assuming that f is continuous on (a,b) and not continuous at x = a.

[tex]\int_{a}^{b} f(x) dx = \lim_{t \to a^+} \int_{t}^{b} f(x) dx = \lim_{t \to a^+} F(x)|_{t}^{b}[/tex]
 
Last edited:
I understand this steps and everything but I still can't see how to get a specific alphas for which the integral converges. How can I find them?
 

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