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I'm sorry to bother you, but I've been studying for a test, and I kinda got stuck in this question. Let me place the question then tell you what I've done so far. I have to find the Fourier Transform of:

$$ f(x)= \begin{Bmatrix} 2a - \left | x \right |, & \left | x \right | < 2a\\ 0, & \left | x \right | > 2a \end{Bmatrix} $$

Then use the result, and Parseval's theorem to evaluate $$ \int_{-\infty }^{\infty} \frac {\sin^{4}(ka)}{k^{4}} dk $$

So far in the first part I split it into the transform of $2a$ minus the transform of $\left | x \right |$, but the result doesn't seem right or of any use the solve the bottom equation. And for the bottom equation I supposed I would have to find the transform of $\sin^{4}(ka) $ times the transform of $\frac {1}{k^4}$

Please help, this is really frustrating :/ Thank you for your time!

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1 Answer 1

The FT of $f$ is

$$\hat{f}(k) = \int_{-2a}^0 dx \: (2 a+x) e^{i k x} + \int_0^{2 a} dx \: (2 a-x) e^{i k x}$$

No way around it - you do out the integrals. By grouping the constant piece and the linear piece in the integrals, you can minimize the algebra. You will cancel out a term of $\sin{2 k a}$ and the result is

$$\hat{f}(k) = \frac{4 \sin^2{a k}}{k^2}$$

To evaluate the integral

$$\int_{-\infty}^{\infty} dk \frac{\sin^4{a k}}{k^4}$$

you use Parseval to see that that integral is equal to

$$16 \int_{-\infty}^{\infty} dk \frac{\sin^4{a k}}{k^4} = 2 \pi \int_{-2 a}^{2 a} dx (2 a-|x|)^2$$

Note that the factor of 16 comes from the factor of $4$ in the FT. The factor of $2 \pi$ on the RHS comes from Parseval's theorem and is on the function side rather than the FT side. The integral on the right is easy when you use the symmetry of the integration interval; the answer is (assuming $a>0$)

$$\int_{-\infty}^{\infty} dk \frac{\sin^4{a k}}{k^4} = \frac{2 \pi}{3} a^3$$

I think your mistake lies in assuming that Parseval states that the integrals of transform pairs are equal. This is false: the integrals of squares of transform pairs are equal, up to a scale factor ($2 \pi$ in this case, depending on how you define the transform).

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Wow this is amazing, thank you so much for this, I really, really appreciate it. I'm working on doing the integrals now and checking that my answer matches the one you said I should get. Also, if you don't mind me asking, what happened to the $\frac {1}{\sqrt {2 \pi}} $ that is usually there for the FTs Thank you so much! –  Angello Jesus Maggio Ferro Mar 18 '13 at 17:59
    
As I said, it depends on the definition of FT you use. You clearly use the symmetric form. I prefer the one where the $1/(2 \pi)$ shows up in the inverse transform only. –  Ron Gordon Mar 18 '13 at 18:02
    
Oh I see, didn't think of that. I did the integrals and checked with mathematica 9 as well, and it worked no problem. Thank you very much for your time and effort! –  Angello Jesus Maggio Ferro Mar 18 '13 at 18:44

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