The interesting thing of the problem is

(Happy)

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- Aug 21st 2009, 03:00 AMsimplependulum
- Aug 21st 2009, 04:57 AMluobo
- Aug 21st 2009, 05:39 AMNonCommAlg
well, i don't know, this might be related to "integration" somehow but this thread is supposed to be about "techniques of integration"! (Nod) so let's just keep it that way please.

i don't want this thread to go in a different direction. cheers! (Smile)

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I'm going to bend the rules a little bit and give you a few integrals instead of one. Pick whichever you like and share your solution with us if you feel like it: (Clapping)

1)

2)

3)

4)

5) (the answer to this one is in terms of Gamma function!)

Have fune! - Aug 21st 2009, 11:22 AMKrizalid
- Aug 21st 2009, 11:36 AMKrizalid
Start with this simple fact: Thus the integral becomes (after reversing integration order) so it remains to compute and its value is which is the answer of the integral. :D

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See below for full answer. I did this when I was in college, so no time to do anything else! - Aug 21st 2009, 02:17 PMNonCommAlg
- Aug 21st 2009, 05:14 PMBruno J.
I attack

First we substitute . Then , i.e. .

Now .

Now consider . Set , so . Then the integral becomes

.

Similarily, consider . Set , so . Then the integral becomes

.

So finally - Aug 21st 2009, 05:22 PMBruno J.
Here is my integral! I posted a particular case of this one not very long ago but here it is anyways because I think it's very nice.

**Determine for which values of****(possibly complex) the following integral exists, and evaluate it for those**:

- Aug 21st 2009, 06:50 PMluobo
- Aug 21st 2009, 07:33 PMluobo
- Aug 21st 2009, 08:01 PMsimplependulum
- Aug 21st 2009, 08:21 PMsimplependulum
- Aug 21st 2009, 08:26 PMluobo
- Aug 22nd 2009, 01:56 AMNonCommAlg
correct!

Quote:

no.4

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Ok, this problem is nice:

__Problem__: Suppose are continuous, is decreasing and is increasing. Prove that for all real numbers

__Note__: Here by we mean The solution I know doesn't use any known integral inequalities. Instead, it uses a nice integration technique and it's fairly short! (Evilgrin) - Aug 22nd 2009, 09:12 AMPaulRS
Since we know that must be non-decreasing - because f is non-increasing- (we can divide by since it is non-zero in our interval), thus we must have

Expand: the inequality will be preserved if we multiply by:

Thus:

Now integrate in and the inequality follows since in -and so the integrals are positive, and we can divide-