Math Challenge Problem:

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- Dec 22nd 2010, 12:00 PMRandom Variabledefinite integral
Math Challenge Problem:

- Dec 22nd 2010, 12:49 PMFernandoRevilla
Let's denote:

with the substitution:

we obtain:

Now, apply twice integration by parts.

Fernando Revilla - Dec 22nd 2010, 01:30 PMRandom Variable
does not have an antiderivative that can be expressed as a finite number of elementary functions. So it's not quite that straightforward.

- Dec 22nd 2010, 01:42 PMchisigma
Setting the integral becomes...

(1)

Now if You consider that is...

(2)

... and...

(3)

... You obtain...

(4)

http://digilander.libero.it/luposaba...ato[1].jpg

Merry Christmas from Italy

- Dec 22nd 2010, 01:54 PMFernandoRevilla
So what?. You can find (definite integral) by means of an equation.

Fernando Revilla - Dec 22nd 2010, 02:08 PMAckbeet
- Dec 22nd 2010, 02:19 PMRandom Variable
- Dec 22nd 2010, 02:24 PMchisigma
Some years ago I 'found' the following formula...

(1)

In that case is so that...

(2)

Thank a lot to Ackbeet!...

http://digilander.libero.it/luposaba...ato[1].jpg

Merry Christmas from Italy

- Dec 22nd 2010, 02:32 PMRandom Variable
My solution:

Let

(since for )

now integrate by parts

(assuming converges uniformly on )

- Dec 22nd 2010, 04:55 PMBruno J.
- Dec 22nd 2010, 11:26 PMFernandoRevilla
**You can**say it. :)

Quote:

but I really don't see how you can find by integration by parts or by "means of an equation" - perhaps you can be more precise and provide your solution?

**I can't**. I had an stupid computational mistake. Sorry. :)

Fernando Revilla - Dec 29th 2010, 09:20 PMTheCoffeeMachine
I'm able to show that:

But I'm lacking the change of variable to show this:

What 'substitution' could do the trick? Mental block!