Comparison Test Question: How to Determine Convergence of a Complex Integral?

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SUMMARY

The discussion focuses on determining the convergence of the integral \(\int_{3}^{\infty } \frac{1}{x + e^x} \,dx\). The user establishes that \(\frac{1}{x + e^x} < \frac{1}{e^x}\) and demonstrates that the integral of \(\frac{1}{e^x}\) converges to \(\frac{1}{e^3}\). The key conclusion is that since \(\frac{1}{x + e^x}\) has less area under its curve compared to \(\frac{1}{e^x}\), it also converges. The user clarifies that knowing \(\frac{1}{x}\) diverges does not imply that \(\frac{1}{x + e^x}\) diverges, as the latter has a smaller area under its curve.

PREREQUISITES
  • Understanding of improper integrals
  • Familiarity with convergence tests for integrals
  • Knowledge of exponential functions and their properties
  • Basic graphing skills to visualize areas under curves
NEXT STEPS
  • Study the comparison test for improper integrals
  • Learn about the behavior of exponential decay in integrals
  • Explore the properties of \(\frac{1}{x^p}\) for different values of \(p\)
  • Investigate other convergence tests, such as the limit comparison test
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Students studying calculus, particularly those focusing on improper integrals and convergence analysis, as well as educators seeking to clarify these concepts for their students.

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Homework Statement



I'm just curious as to how to think about the following form of equation.

Homework Equations


[tex]\int_{3}^{\infty } \frac{1}{x + e^x} \,dx[/tex]


The Attempt at a Solution



What you're trying to do is to test it;

[tex]\frac{1}{x \ + \ e^x} \ < \ \frac{1}{x}[/tex]

[tex]\frac{1}{x}[/tex] diverges

[tex]\frac{1}{x \ + \ e^x} \ < \ \frac{1}{e^x}[/tex]

[tex]\lim_{t \to \infty} \int_{3}^{t} e^{-x}\,dx \ = \ \lim_{t \to \infty} - e^{-x} | \ _3 ^t \ = \ \lim_{t \to \infty} - {\frac{1}{e^t} \ + \ \frac{1}{e^3}[/tex]

so this converges to [tex]\frac{1}{e^3}[/tex].

I don't get how this means the original eq. will also converge?

Both [tex]\frac{1}{x}[/tex] and [tex]\frac{1}{e^x}[/tex] are bigger than the original eq. with one converging and the other diverging.

[tex]\frac{1}{x}[/tex] is bigger than [tex]\frac{1}{e^x}[/tex] but the test of [tex]\frac{1}{x^p}[/tex] is ringing in my ears as a kind of explanation, but I am confused.

Is there an easy way to link all of this together?
 
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Indeed, because for [strike]positive x[/strike] x > 1 when p > 1, 1/xp is always less than 1/x. In effect, its graph has less area underneath it for the integration. So there is no contradiction that 1/x leaves just too much area, i.e. it diverges, while 1/xp for p > 1 does not, i.e. it converges. Meanwhile, 1/ex just hugs the x-axis massively more than 1/xp for any constant p, hence it of course converges. Finally, 1/(x+ex) lies even closer to the x-axis and leaves even less area; it converges.

Knowing 1/x diverges does not mean we know 1/(x+ex) diverges or converges. The graph of the latter has less area, but by how much? Enough to stop the integral from diverging? However, knowing that 1/ex converges does tell us that 1/(x+ex) converges, since it has less area than a finite amount.

EDIT: It occurred to me that I also need x > 1, not just x positive. Sorry for the mistake.
 
Last edited:
Ahh Brilliant!

I think that idea of hugging the x-axis makes everything even clearer, thanks a lot for that :wink:
 

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