- #1
RedX
- 970
- 3
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.
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.