# Fourier transform

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:

$$\int^{\infty}_{-\infty} \frac{dx}{x} e^{-ikx}$$

not converge because of x=0?

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

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

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

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?

$$f(x)=\int \frac{dk}{2 \pi} (i \pi) e^{-ikx}=(i \pi)\delta(x)\neq1/x$$

AlephZero
Homework Helper
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.

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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.:

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

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

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