Fourier Transform of x^2*sin(x)

Printable View

Show 40 post(s) from this thread on one page

July 22nd 2010, 06:30 AM
lvleph

Fourier Transform of x^2*sin(x)

First let me define the Fourier Transform of f to be

$\mathcal{F}_x[f](\lambda) = \int_{-\infty}^{+\infty}\! f(x) e^{ix\lambda}\,dx$

I am trying to transform $x^2\cdot sin(x)$ , 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 AM
Ackbeet

The $x^{2}$ 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 AM
lvleph

Quote:

Originally Posted by Ackbeet

The $x^{2}$ 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.

Thank you, but I cannot use a table. I actually have to integrate this.
July 22nd 2010, 06:59 AM
Ackbeet

Ok. Try by-parts, with $u$ being $x^{2}$ , and $dv=\sin(x)e^{ix\lambda}.$ Integrating your $dv$ 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 AM
lvleph

Let's see if I am getting this right
Let $u = x^2 \Rightarrow \frac{du}{dx} = 2x$ ,
$dv = \sin x \cdot e^{ix\lambda} \Rightarrow v = \frac{1}{i\lambda} \left[\sin x \cdot e^{ix\lambda} - \int_{-\infty}^{+\infty}\! \cos x \cdot e^{ix\lambda}\, dx\right]$
$\int_{-\infty}^{+\infty}\! u\cdot dv\, dx = \left.u \cdot v\right|_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty}\! du\cdot v\, dx$
and so we have
$\int_{-\infty}^{+\infty}\! x^2\cdot \sin x \cdot e^{ix\lambda}\, dx$
$= \left.\frac{x^2}{i\lambda} \left(\sin x \cdot e^{ix\lambda} - \int_{-\infty}^{+\infty}\! \cos x \cdot e^{ix\lambda}\, dx\right)\right|_{-\infty}^{+\infty} - \frac{2}{i\lambda}\int_{-\infty}^{+\infty}\! x\cdot \left(\sin x \cdot e^{ix\lambda} - \int_{-\infty}^{+\infty}\! \cos x \cdot e^{ix\lambda}\, dx\right)\, dx$
July 22nd 2010, 07:48 AM
Ackbeet

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 AM
lvleph

I was just checking my steps.
$\int\! \sin x \cdot e^{ix\lambda}\, dx = \frac{e^{ix\lambda}}{1 - \lambda^2}\left(i\lambda \sin x - \cos x\right)$
Now I just have to do the next steps. Thank you for your help.
July 22nd 2010, 07:58 AM
Ackbeet

Correct. Now just turn the crank.
July 22nd 2010, 08:00 AM
lvleph

So the next problem is that $\int_{-\infty}^{+\infty}\! \sin x\cdot e^{ix\lambda}\, dx$ is not convergent.
July 22nd 2010, 08:16 AM
Ackbeet

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 AM
CaptainBlack

Quote:

Originally Posted by lvleph

First let me define the Fourier Transform of f to be

$\mathcal{F}_x[f](\lambda) = \int_{-\infty}^{+\infty}\! f(x) e^{ix\lambda}\,dx$

I am trying to transform $x^2\cdot sin(x)$ , but I haven't a clue where to begin here. I am sure that I am just not seeing some trick. Thanks in advance.

You can't define this FT that way since the integral does not converge for any real $$$ \lambda$ , it is the analog of asking for the sum:

$\displaystyle \sum_{-\infty}^{\infty} (-1)^n n^2$

Your function does have a distribution as its FT

CB
July 22nd 2010, 11:50 AM
Ackbeet

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:

$\displaystyle{i\sqrt{\frac{\pi}{2}}(\delta''(t-1)-\delta''(t+1))}.$
July 22nd 2010, 11:57 AM
CaptainBlack

Quote:

Originally Posted by Ackbeet

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:

$\displaystyle{i\sqrt{\frac{\pi}{2}}(\delta''(t-1)-\delta''(t+1))}.$

Yes I know it has a FT as a distribution, but the OP has excluded that approach.

CB
July 22nd 2010, 11:59 AM
Ackbeet

Quote:

...the OP has excluded that approach

I know I'm showing my ignorance of distribution theory here (which is no more than the truth), but how has the OP excluded distributions from the allowed solutions?
July 22nd 2010, 12:09 PM
CaptainBlack

Quote:

Originally Posted by Ackbeet

I know I'm showing my ignorance of distribution theory here (which is no more than the truth), but how has the OP excluded distributions from the allowed solutions?

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

Show 40 post(s) from this thread on one page