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I would like to know which is the fourier transform of integral operator:

$$Tf(x)=\int_{-\infty}^{+\infty}\quad f(x)dx\rightarrow \hat{T}\hat{f}(k)$$

I know that (is it right?):

$$\int_{-\infty}^{+\infty}\quad x^2f(x)dx\rightarrow -\frac{\partial^2}{\partial k^2}\hat{f}(k)|_{k=0}$$

Thank you

JFNJr
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1 Answers1

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You seem to be mistaken here. The Fourier transform is a integral operator (when acting on certain function spaces). It takes a function and gives another function. So $\mathcal{F}:f\rightarrow \hat{f}$. What you have written here doesn't make sense because you are asking for the Fourier transform of a constant (which makes sense distributionally but not in the classical sense) since definite integrals are just constants. Additionally, all information is lost about $f$ in what you have written. What you want to say is that the Fourier transform maps differentiation to multiplication and vice versa which is true. To be more precise:

$$ \int_{\mathbb{R}} \frac{e^{-ikx}}{\sqrt{2\pi}} x^2f(x)dx = -\frac{d^2\hat{f}}{dk^2} $$

Cameron Williams
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  • No, I think he is saying that multiplication by $x^2$ in an integrand is the same as evaluating the 2nd derivative of the FT at $k=0$. It is not claiming to take the FT of a constant. – Ron Gordon May 14 '13 at 13:35
  • The expressions he wrote are entirely meaningless and suggest a fundamental misunderstanding which I was trying to correct. – Cameron Williams May 14 '13 at 13:38
  • The 2nd equation is entirely correct as far as I can see, so long as the FT is defined in terms of $k$, etc., in the usual way. The first equation is, to be generous, kind of lacking. – Ron Gordon May 14 '13 at 13:39
  • Neither expression is right. The Fourier transform takes a function $f$ not its definite integral because that is the same as just taking the transform of a constant and all information about $f$ is lost. OP seems to think that $\mathcal{F}\left(\int_{\mathbb{R}} f(x)dx\right)$ gives some meaningful expression. – Cameron Williams May 14 '13 at 13:47
  • $$\begin{align}\frac{d^2}{dk^2} \hat{f}(k) &= \frac{d^2}{dk^2} \int_{-\infty}^{\infty} dx , f(x) , e^{i k x} \ &= -\int_{-\infty}^{\infty} dx , x^2 , f(x), e^{i k x} \end{align}$$

    Therefore

    $$\left[ \frac{d^2}{dk^2} \hat{f}(k)\right]{k=0} = -\int{-\infty}^{\infty} dx , x^2 , f(x)$$

     – Ron Gordon May 14 '13 at 13:50
  • I don't argue that what you are saying is correct but what I am saying is that what OP has written is meaningless. The Fourier kernel is missing and OP thinks the Fourier transform acts on integral operators (which is NOT what he wants.. hence I think OP is mistaken). – Cameron Williams May 14 '13 at 13:54
  • Actually I think OP is slightly confused or using really bad notation but the answer is right. The arrow suggested he was taking a Fourier transform of that data instead of having equality like he should. – Cameron Williams May 14 '13 at 13:57
  • I will agree with that last sentiment. – Ron Gordon May 14 '13 at 13:57
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    The way OP wrote it was massively confusing but I think you interpreted what OP meant correctly. OP should have written equality rather than $\rightarrow$. – Cameron Williams May 14 '13 at 13:59