Convergence of an Improper Integral Involving Exponential Functions

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SUMMARY

The discussion focuses on determining the convergence of the improper integral \(\int_{0}^{\infty} \frac{dx}{e^{-x} + e^{x}}\). The initial substitution \(u = e^x\) leads to the transformed integral \(\int_{0}^{\infty} \frac{u \, du}{u^{2} + 1}\). However, the approach was criticized for inaccuracies, particularly regarding limits and integration steps. Participants emphasized the importance of a meticulous, step-by-step method to ensure correct evaluation of the integral.

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  • Understanding of improper integrals
  • Familiarity with exponential functions
  • Knowledge of substitution methods in integration
  • Basic skills in evaluating limits
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  • Study techniques for evaluating improper integrals
  • Learn about convergence tests for integrals
  • Explore integration by substitution in detail
  • Review the properties of exponential functions and their integrals
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Students studying calculus, particularly those focusing on integration techniques and improper integrals, as well as educators looking for examples of common pitfalls in integral evaluation.

SirPlus
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Homework Statement



1.Determine the divergence/convergence of the following improper integrals by the evaluation of the limit:

\int_{0}^{∞} \frac{dx}{e^{-x} + e^{x}}




Homework Equations





The Attempt at a Solution



Let u = e^x
∴ du = e^x dx

I ended up with:

\int_{0}^{∞} \frac{u du}{u^{2} + 1}
I have no idea on how to integrate the above ...
 
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substitute v=u^2.
 
Hi SirPlus! :smile:
SirPlus said:
\int_{0}^{∞} \frac{dx}{e^{-x} + e^{x}}

Let u = e^x
∴ du = e^x dx

fine so far :wink:
\int_{0}^{∞} \frac{u du}{u^{2} + 1}

Sorry, but that's completely wrong (including the limits).

Start again, and this time write it out carefully step-by-step, with no short-cuts! :smile:
 

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