This one was given to me as a challenge. I was happy I could solve it!

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- Aug 3rd 2009, 07:04 PMBruno J.Challenge integral
This one was given to me as a challenge. I was happy I could solve it!

Show that

- Aug 3rd 2009, 09:01 PMKrizalid
It's quite easy.

Hence we get as required. - Aug 3rd 2009, 09:02 PMTwistedOne151
First, use integration by parts with , , so that , , and

Now, let : x=0 becomes , x=1 becomes , and , and , so our right-hand integral becomes

.

Now,

(see here).

So

.

-Kevin C. - Aug 3rd 2009, 09:31 PMRandom VariableQuote:

I don't understand what you did here. - Aug 3rd 2009, 09:32 PMKrizalid
and then switch sum and integral.

- Aug 4th 2009, 11:28 AMMoo
- Aug 4th 2009, 01:52 PMKrizalid
you know the justification, and it's quite known and I won't do it.