Posted by T.Gracken on October 15, 2002 at 16:37:00:
In Reply to: not the T.Gracken that posted earlier. posted by T.Gracken on October 15, 2002 at 16:12:00:
: : improper integral of 1/x^2 from 0 to 1
: : Doesn't exist, correct???
you are correct.
to show it...
integral (from 0 to 1) of {1/x2}dx
= lim (k -> 0+) of [integral (from k to 1) of {1/x2}dx
= lim (k -> 0+) of [-1/x](evaluated from k to 1)
= lim (k -> 0+) [-1 + 1/k]
= 1 + infinity (i.e. diverges)
therefor the integral can not be evaluated.