Article View: k12.ed.math
Article #496Re: INTEGRALS
From: David G Radcliffe
Date: Thu, 21 Jan 1993 17:19
Date: Thu, 21 Jan 1993 17:19
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968 bytes
I think some people are making this more complicated than it needs
to be. By definition, I{-2,2} 1/x = I{-2,0} 1/x + I{0,2} 1/x
*provided* that the integrals on the right converge. Both of the
integrals on the right diverge, therefore the integral on the left
diverges. If you look at the *definition* of an improper integral,
rather than doing a lot of intuitive handwaving, then everything
becomes clear.
--
David Radcliffe
radcliff@csd4.csd.uwm.edu
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