Fourier transform

  • Thread starter RedX
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  • #1
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Does it make sense to take the Fourier transform of a function that blows up at some point? For example the Fourier transform of f(x)=1/x, which blows up at zero?

Doesn't the integral:

[tex]\int^{\infty}_{-\infty} \frac{dx}{x} e^{-ikx} [/tex]

not converge because of x=0?

Yet for some reason analytical computer programs give a result despite the blow up.
 

Answers and Replies

  • #2
2,967
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[tex]
P \int_{-\infty}^{\infty}{\frac{e^{i k x}}{x} \, dx} = i \pi
[/tex]
 
  • #3
970
3
[tex]
P \int_{-\infty}^{\infty}{\frac{e^{i k x}}{x} \, dx} = i \pi
[/tex]
Well if the Fourier transform of 1/x exists, it would make the most sense that it's the principal part.

However, can you really ignore what's going on at x=0 like that?

For example, can you reconstruct the function from the transform?

[tex]f(x)=\int \frac{dk}{2 \pi} (i \pi) e^{-ikx}=(i \pi)\delta(x)\neq1/x [/tex]
 
  • #4
AlephZero
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This is rather a subtle question. The answer probably depends on what you want to do with the FT when you have got it.

The FT is "defined", in the sense that sin(kx)/x is a well behaved integrable function everywhere except at when x = 0, and cos(kx)/x is an odd function so in some sense the integral from -infinity to +infinity must be 0, even though the integrals from -infinity to 0 and 0 to +infinity both diverge.

But as RedX implied, "here be mathematical dragons..." - though this sort of mathematical arm-waving is sometimes useful for discovering correct and useful results, which can be proved by other methods once they are known.
 
Last edited:
  • #5
970
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I made a mistake. The Fourier transform of 1/x actually has a sign function in it:

http://www.wolframalpha.com/input/?i=fourier+1/x

because depending on whether k in the exponential is positive or negative, you have to complete your contour in the upper or lower half respectively of the complex x-plane, i.e.:

[tex]

P \int_{-\infty}^{\infty}{\frac{e^{i k x}}{x} \, dx} = i \pi

[/tex]


[tex]

P \int_{-\infty}^{\infty}{\frac{e^{-i k x}}{x} \, dx} = -i \pi

[/tex]


So maybe there's a chance that 1/x is the inverse Fourier transform.
 

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