Find where an improper integral converges

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 1K views
rocapp
Messages
95
Reaction score
0

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?
 
Physics news on Phys.org
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.