What is the Limit of the Natural Logarithm of Infinity Minus One?

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Discussion Overview

The discussion revolves around the evaluation of the limit of the natural logarithm of infinity minus one, specifically in the context of the integral \(\int^{\infty}_{1}\frac{1}{e^{t}-1}dt\). Participants explore the behavior of this integral as \(t\) approaches infinity and the implications for the limit of the logarithmic expression.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about how the integral evaluates to \(-\ln(e - 1) + 1\) and questions how the \(+1\) arises from the limit as \(t\) approaches infinity, suggesting it should yield infinity instead.
  • Another participant asks for clarification on the anti-derivative found and inquires about the limit as \(t\) approaches infinity.
  • A third participant proposes a substitution \(e^t=u\) to transform the integral, leading to a new expression involving logarithms, which they claim evaluates to \(1 - \log(e - 1)\) after considering the limit at infinity.
  • A later reply reiterates the original integral evaluation and suggests that the limit can be expressed as \(\lim\limits_{a \rightarrow \infty} (-\ln|a| + \ln|a-1| + ...)\), prompting further exploration of how to combine the logarithmic terms for evaluation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the evaluation of the integral or the limit. There are competing views on how to approach the problem and differing interpretations of the behavior of the logarithmic expression as \(t\) approaches infinity.

Contextual Notes

There are unresolved assumptions regarding the convergence of the integral and the behavior of the logarithmic terms at infinity. The discussion reflects varying levels of understanding and approaches to the problem without definitive conclusions.

autobot.d
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[itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}-1}dt[/itex]

[itex]= -ln(e - 1) + 1[/itex]

Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e.
[itex]ln(e^{\infty} -1) = ?[/itex]


Any help appreciated, thanks.
 
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What anti-derivative did you find?
What is the limit as [itex]t \to \infty[/itex]?
 
autobot.d said:
[itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}-1}dt[/itex]

[itex]= -ln(e - 1) + 1[/itex]

Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e.
[itex]ln(e^{\infty} -1) = ?[/itex]


Any help appreciated, thanks.



I don't know how you solve this integral, but I must make a substitution:

[itex]e^t=u\Longrightarrow t=\log u \Longrightarrow dt=\frac{du}{u}[/itex] , so the integral becomes:

[itex]\int^{\infty}_e \frac{du}{u(u-1)}=\int^\infty_e\frac{du}{u-1}-\int^\infty_e\frac{du}{u}= 1-\log (e-1)[/itex] , after evaluating the limit in infinity...

DonAntonio
 
autobot.d said:
[itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}-1}dt[/itex]

[itex]= -ln(e - 1) + 1[/itex]

Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e.
[itex]ln(e^{\infty} -1) = ?[/itex]


Any help appreciated, thanks.

Eventually your answer will boil down to [itex]\lim\limits_{a \rightarrow \infty} (-ln|a| + ln|a-1| + ...)[/itex] (as you can see from DonAntonio's work above).

You need to take this limit, can you think of a way to combine the two natural logarithms in order to do this?
 

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