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Consider the function in $\mathbb{R}^3$ given by $f(\vec{x})=\frac{1}{x}$ where $x$ is the modulus of the point $\vec{x}$. It is a well known fact that the Fourier transform of this function

$$\tilde{f}(\vec{q})=\int{}d^3x\,e^{-i\vec{q}\vec{x}}\frac{1}{x}=\frac{4\pi}{q^2}$$

Nonetheless, $f$ is ill defined at the point $0$. The integral aboive goes through zero, how is this integral even defined?

PhoenixPerson
  • 717
  • $f$ is not an integrable function, so you need to use a different definition of the Fourier transform anyway. Since the singularity at $0$ is integrable, $f$ defines a tempered distribution. Compute the Fourier transform of this tempered distribution. – Daniel Fischer Mar 03 '16 at 13:50

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