🚀 go-pugleaf

In article <40110.2B5EBAC2@puddle.fidonet.org>, Sheila.King@f315.n103.z1.fidonet.org (Sheila King) writes:
|> Hi Henry Choy,
|> 
|> Reading your message in response to Mike Bouldan, regarding integrating
|> 1/x from -2 to 2 you say:
|> 
|> -> Just integrate from -2 to 0 (NOT FROM 0 to -2) and then from 0 to 2.
|> -> You have to take the limits each time. But what is -infinity +
|> -> infinity? You don't know, so you add the positive and negative halves
|> -> of the series together: i.e. I{0,2}(1/x-1/x)dx = 0 in the limit.
|> 
|> That the answer is zero. Which seems to make sense to me (not that I'm
|> any authority). However, Mike didn't think this was the right answer,
|> and it doesn't agree with Brian Harvey's post, so I'm a little puzzled.
|> Are you _positive_ this is the answer to this problem?

Sorry. I made a mistake. There is no answer. The limit is undefined
because the left and right side limits don't agree (+ and - infinity).
See earlier post.

You can't even use L'Hopital's rule, which is a rule proven for
normal limits, not the Reimann sums used in definite integrals.

-- 

Henry Choy
choy@cs.usask.ca

What rolls down stairs alone or in pairs            This has been brought to 
Rolls over your neighbor's dog?                     you by the numbers 4
What's great for a snack and fits on your back?     and 9 and the letter P.
It's Log, Log, Log!  -- "The Log Song", from            -- Big Bird
                         Ren & Stimpy

Math is tough!       -- Barbie