Compute

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- May 11th 2010, 09:10 PMDrexel28Easy integral?
Compute

- May 11th 2010, 11:46 PMsimplependulum
- May 12th 2010, 03:48 AMtonio
- May 12th 2010, 05:30 AMsimplependulum
I write

then substitute

At first i did it by integration by parts but the steps are quite long so i was not sure at that time ... Now , i am sure it is true with your help .(Happy) - May 12th 2010, 02:21 PMDrexel28
Just tinkering

There is another method that involves complex analysis, and it is the way I was hoping someone would do this.

We know that and thus , in particular the real an imaginary parts of that integral must equal zero. Do a little work with trig identities and you will get our integral is the imaginary part of it. - May 12th 2010, 04:32 PMRandom Variable
let

then

let

EDIT: then

so

but

which means

so

and