Posted by flo on October 15, 2002 at 16:39:35:
In Reply to: here's one way (hopefully this will shut the fake T.G. up) posted by T.Gracken on October 15, 2002 at 16:37:00:
: : : 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.