Improper integral convergence or divergence.

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SUMMARY

The discussion focuses on determining the convergence or divergence of the improper integral from 0 to infinity of arctan(x) / (2 + e^x) using the Comparison Theorem. Participants noted that comparing this integral to 1/e^x was ineffective, as the latter converges while the former is less than it. A key insight is that the integral's convergence is assured if it is less than a known convergent integral, rather than greater.

PREREQUISITES
  • Understanding of improper integrals
  • Familiarity with the Comparison Theorem
  • Knowledge of the behavior of arctan(x) as x approaches infinity
  • Basic calculus concepts related to convergence tests
NEXT STEPS
  • Study the Comparison Theorem in detail
  • Explore examples of improper integrals and their convergence
  • Learn about the behavior of arctan(x) for large values of x
  • Investigate other convergence tests such as the Limit Comparison Test
USEFUL FOR

Students and educators in calculus, particularly those focused on improper integrals and convergence analysis, as well as anyone looking to deepen their understanding of the Comparison Theorem.

crazy_nuttie
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Homework Statement



Use Comparison Theorem to determine whether the integral is convergent or divergent:

integral from 0 to infinity of: arctan(x) / (2 + e^x)

Should look like this: http://bit.ly/cAhytV

Homework Equations



--

The Attempt at a Solution




I tried to compare this with the integral from 0 to infinity of 1/e^x, but that didnt succeed since 1/e^x converges and the given function is less than 1/e^x. I am not sure what to compare this with
 
Last edited by a moderator:
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crazy_nuttie said:

Homework Statement



Use Comparison Theorem to determine whether the integral is convergent or divergent:

integral from 0 to infinity of: arctan(x) / (2 + e^x)

Should look like this: http://bit.ly/cAhytV

Homework Equations



--

The Attempt at a Solution




I tried to compare this with the integral from 0 to infinity of 1/e^x, but that didnt succeed since 1/e^x converges and the given function is less than 1/e^x. I am not sure what to compare this with

Not sure why you chose 1/ex since \arctan x \le \pi/2 but not bounded by 1.

Anyway, why wouldn't you have succeeded if your integral is less than a known convergent integral? Your problem would be if your unknown integral was greater than a known convergent integral.
 
Last edited by a moderator:

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