Comparing and Checking Improper Integrals

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Discussion Overview

The discussion revolves around the evaluation of an improper integral, specifically ∫(2 to ∞) dv/(v^2 + 2v - 3). Participants are checking their answers and exploring methods for verifying their results, including the handling of limits at infinity and the use of logarithmic properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their computed answer as (1/4)ln(1/5) and seeks comparison with others' results.
  • Another participant questions the handling of limits at infinity and points out a discrepancy in signs between the computed answers.
  • A participant agrees that their answer should have been negative after further evaluation of logarithmic expressions.
  • One participant reports a different result of -(1/4)ln(3) after integrating and recombining logarithms.
  • Another participant expresses confusion regarding the appearance of ln(3) and provides their limit evaluation leading to -1/4ln(1/5).

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as multiple competing views and results are presented regarding the evaluation of the integral and the handling of logarithmic expressions.

Contextual Notes

There are unresolved issues regarding the correct handling of limits at infinity and the application of logarithmic properties, which may affect the final results. The discussion reflects varying interpretations of the integral's evaluation process.

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Hi guys just want to check my answer for the following improper integral.
∫(2 to ∞) dv/v^2+2v-3. After doing partial fractions, integrating and evaluating I got the following for the answer: 0-(1/4)ln(1/5)=(1/4)ln(1/5)
How does this compare to other answers?
Is there a way I can accurately check this answer myself?
 
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1) How are you handling the limits at infinity?

2) Certainly, -(1/4)ln(1/5) is not equal to (1/4)ln(1/5)!
 
I made the natural logs into a quotient by log rules and evaluated that for t approaching infinity. I do agree with you my answer should have been negative.
 
That's not what I get. After integrating and recombining the logarithms, I get -(1/4)ln(3) at infinity.
 
I'm not sure how you got ln of 3 before evaluating I have lim t-> infinity 1/4ln(v-1/v+3) evaluated from 2 to t.
Lim t->infinity [1/4ln((t-1)/(t+3))-1/4ln((2-1)/2+3))
Giving me -1/4ln(1/5)
 

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