Find where an improper integral converges

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SUMMARY

The improper integral ∫-∞∞(dx/x²) diverges due to the behavior of the function 1/x² as x approaches 0. The integral must be split into two parts: ∫(-∞ to 0) dx/x² and ∫(0 to ∞) dx/x². Both of these integrals diverge to infinity, confirming that the overall integral does not converge. The key takeaway is the importance of recognizing discontinuities in the function when evaluating improper integrals.

PREREQUISITES
  • Understanding of improper integrals
  • Knowledge of limits and continuity in calculus
  • Familiarity with the concept of divergence in integrals
  • Basic integration techniques, particularly for rational functions
NEXT STEPS
  • Study the properties of improper integrals in calculus
  • Learn about convergence tests for integrals
  • Explore the concept of discontinuities in functions
  • Review integration techniques for rational functions, focusing on limits
USEFUL FOR

Students studying calculus, particularly those focusing on integral calculus and improper integrals, as well as educators teaching these concepts.

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Homework Statement



-∞(dx/x2)

Homework Equations





The Attempt at a Solution



∫(dx/x2) = -1/x

(-1/∞) - (-1/-∞) = 0

However, the answer is that the integral diverges. Why is this the case?
 
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The function [itex]1/x^2[/itex] grows infinitely large as [itex]x \rightarrow 0[/itex], so you have to break this into two improper integrals:
[tex]\int_{-\infty}^{0} dx/x^2 + \int_{0}^{\infty} dx/x^2[/tex]
You can easily check that both of these integrals diverge to [itex]\infty[/itex].
 
Thanks! I forgot that it was dependent on the function being continuous.
 

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