Most of us know that

But what about ?

The worked solution I've seen is quite confusing.

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- April 18th 2010, 09:23 PMRandom Variableanother challenging integral
Most of us know that

But what about ?

The worked solution I've seen is quite confusing. - April 18th 2010, 10:03 PMPulock2009by parts integration
- April 19th 2010, 02:20 AMRandom Variable
- April 19th 2010, 02:45 AMsimplependulumOpinion
I convert this integral to another (unknown) one , not sure if it is useful .

I consider the integral

Since and

I find that

It is still an unknown integral but I can represent it by infinite series .

I am curious if this integral can be expressed by somthing ... - April 19th 2010, 03:46 AMRandom Variable
You end up finding the value of both integrals.

Let

and

You should find that

and (I'm a bit confused by this part.)

and then just solve simultaneously for and - April 19th 2010, 04:49 AMsimplependulum
I got it !

I first introduce two lemmas :

The proofs are left to you , they should be easy . Note that the first one might be proved using Fourier Analysis, I think .

Use

For single-valued function , we have

So we have ,

Sub for the second integral ,

Consider the integral

Sub. in the second one , we can see that the integral equals to :

Also from my previous post , we obtain ( use your symbols )

Summation gives

We also obtain a by-product

Just after the last line , I thought can we generalize the integral , power up to n ? I really want to go through it but i won't have much time since i am busy on my school examination (Crying) . - April 19th 2010, 08:24 AMRandom Variable
Thanks. They just stated that without any explanation. The overall proof that they give is slightly different than yours, but not different enough to state it here.

Anyways, you can prove by doing the following:

(let )

( let )

- April 19th 2010, 08:34 AMRandom Variable
I can't think of the the appropriate Fourier series for the first lemma.

- April 19th 2010, 01:53 PMRandom Variable
I found a proof, but it's a bit convoluted.

The Fourier series of on is

now integrate both sides from 0 to x

so

now let

then - April 19th 2010, 10:30 PMsimplependulum

The proof by expanding Fourier series I think is too general and a bit boring , also hard to memorize . I know Euler gave a general but not boring method to solve this kind of series but i forgot it ( it is hard to memorize too (Happy) ) . Once i find out the method , i post it here immediately , i think we may find it inspiring . - April 20th 2010, 12:10 AMRandom Variable
If you're still looking for , Maple gives the following results: