Here's one to give a go if you so desire. We always like a new integral to chew on, huh?. (Cool)(Happy)

Evaluate:

I know, I know, it's easy. Give it a go anyway.(Rock)

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- Dec 26th 2008, 01:29 PMgalactusintegral challenge
Here's one to give a go if you so desire. We always like a new integral to chew on, huh?. (Cool)(Happy)

Evaluate:

I know, I know, it's easy. Give it a go anyway.(Rock) - Dec 26th 2008, 02:01 PMNonCommAlg
- Dec 26th 2008, 02:17 PMgalactus
You more than likely went about it easier than I.

By letting

I left

Then setting

Assuming they exist of course.

Anyway, it took me longer than it did you, NCA. You are one sharp cat. I always like to peruse your posts for something useful and new to learn.

You've forgotten more than I know.(Bow)(Giggle) - Dec 26th 2008, 02:22 PMJester
- Dec 26th 2008, 03:22 PMNonCommAlg
a shorter way is to notice that your integral is a Frullani's Integral if you let

- Dec 26th 2008, 03:30 PMgalactus
Cool Observation. I must admit, I should have saw that.

- Dec 26th 2008, 03:32 PMMathstud28
Besides the above you could see right away that we can discompose our integral as follows

So our integral becomes

Now using Frullani's Theorem which states that assuming that

So if we let

We get

EDIT: NonCommAlg...you ruin all my fun. ;) - Dec 26th 2008, 03:33 PMJester
I once saw the following on a Putnam exam

What a wonderful generalization in the link you provide - thanks for sharing! (Clapping) - Dec 26th 2008, 03:35 PMMathstud28
- Dec 26th 2008, 03:52 PMNonCommAlg
- Dec 26th 2008, 03:57 PMMathstud28
- Dec 26th 2008, 04:01 PMJester
- Dec 26th 2008, 04:06 PMNonCommAlg
- Dec 26th 2008, 04:10 PMMathstud28
- Dec 26th 2008, 04:11 PMJester