First let me define the Fourier Transform of f to be

I am trying to transform , but I haven't a clue where to begin here. I am sure that I am just not seeing some trick. Thanks in advance.

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- July 22nd 2010, 06:30 AMlvlephFourier Transform of x^2*sin(x)
First let me define the Fourier Transform of f to be

I am trying to transform , but I haven't a clue where to begin here. I am sure that I am just not seeing some trick. Thanks in advance. - July 22nd 2010, 06:52 AMAckbeet
The term you can handle by using line 107 here. The sin function you can handle by line 305 on the same page. It's not pretty, but you can get a closed-form expression for your function, despite not being square-integrable.

- July 22nd 2010, 06:55 AMlvleph
- July 22nd 2010, 06:59 AMAckbeet
Ok. Try by-parts, with being , and Integrating your will itself require by-parts. Then you'll have to do the process all over again. That should get rid of the powers of x out front. What do you get after that?

- July 22nd 2010, 07:44 AMlvleph
Let's see if I am getting this right

Let ,

and so we have

- July 22nd 2010, 07:48 AMAckbeet
Hmm. Check your dv integration. You have to do by-parts twice on functions like that in order to get back to your original integral. You then pull it over to the other side and solve for it. You follow?

[EDIT]: Also, check whether you're computing definite or indefinite integrals, and where you're doing that. Don't mix them up! - July 22nd 2010, 07:56 AMlvleph
I was just checking my steps.

Now I just have to do the next steps. Thank you for your help. - July 22nd 2010, 07:58 AMAckbeet
Correct. Now just turn the crank.

- July 22nd 2010, 08:00 AMlvleph
So the next problem is that is not convergent.

- July 22nd 2010, 08:16 AMAckbeet
Hmm. Are you sure you need to compute that? Finish computing the integral of dv. Then write out your outer by-parts, and type that up. Let me see what you have there.

- July 22nd 2010, 11:42 AMCaptainBlack
- July 22nd 2010, 11:50 AMAckbeet
Reply to CB at Post # 11:

Well, I was hoping that through the computations being done above, we'd be able to get to the point of recognizing the integrals that are left as the second derivatives of the Dirac Delta function. There is an inverse FT for that function:

__Spoiler__: - July 22nd 2010, 11:57 AMCaptainBlack
- July 22nd 2010, 11:59 AMAckbeetQuote:

...the OP has excluded that approach

- July 22nd 2010, 12:09 PMCaptainBlack
Post #3. You need a non-standard definition of an integral to do this as an intergral, and I'm pretty sure the OP has never seen such a definition (of course the OP may be an engineering student, in which case the lecturer may just wave their hands over such things and >POOFF< there it is).

CB