Okay, I have a fairly stupid question.

How do I check wheter integral

exists?

When I have

, I have to check that f is bounded and continuous almost everywhere. Do I have to do the same when I have ? Is it enough to do the same?

Thank you.

Printable View

- Oct 24th 2008, 03:34 PMmarianneChecking whether integral exists
Okay, I have a fairly stupid question.

How do I check wheter integral

exists?

When I have

, I have to check that f is bounded and continuous almost everywhere. Do I have to do the same when I have ? Is it enough to do the same?

Thank you. - Oct 24th 2008, 04:15 PMJameson
If you are computing , then the additional requirement of existence is that the following two limits exist and are finite: and , where c is some constant that is convenient for your function.

(Summed up from Wikipedia article on Improper Integrals) - Oct 24th 2008, 04:17 PMgalactus

If either integral on the right side diverges, then we say that

diverges.

Let f be continuous on some interval, say, [a,b]. With the exception of at

some point c satisfying a<c<b. f(x) becomes infinite as x approaches c from

the left or right. If the two improper integrals

or

both converge, then we say that the improper integral

converges. That way we define:

See what I mean?. Is that what you were getting at?.

You can also say

The thing is, check for convergence or divergence of the limit.

Take for instance.

The other side can be shown to be the same and we have as the solution.

But, we split the integral at x=0. We did not have to do that. We could have done it anywhere and not affected the convergence or divergence.

Does that help a wee bit?. - Oct 24th 2008, 04:19 PMtoraj58

is equal with:

- Oct 25th 2008, 01:25 AMmarianne
Thank you all, you've been most helpful!

I think I got it. :-) The example galactus provided really sorted it out for me.