$$ f(x) = \frac{1}{1+ix} $$

in a textbook I am told that its Fourier transform is:

$$ \hat{\omega} = 2\pi~\Theta(-\omega)~e^\omega $$

I am aware of the relation between the Heaviside step function and the Dirac delta function, however by applying the definition of a Fourier transform I am not able to derive the above result.


Correction: It must be $\dfrac{1}{1+i\omega}$ not $\dfrac{1}{1-i\omega}$.

First we try to find the Fourier transform of $g(x)=e^{-x}\Theta(x)$$$G(\omega)=\int_{-\infty}^{\infty}e^{-x}\Theta(x)e^{-i\omega x}dx=\int_{0}^{\infty}e^{-x}e^{-i\omega x}dx=\dfrac{e^{-x(1+i\omega)}}{1+i\omega}|_{\infty}^{0}=\dfrac{1}{1+i\omega}$$so$$g(x)\Leftarrow\Rightarrow \dfrac{1}{1+i\omega}$$by using duality:$$\dfrac{1}{1+it}\Leftarrow\Rightarrow2\pi g(-\omega)$$and$$\dfrac{1}{1+it}\Leftarrow\Rightarrow2\pi e^{\omega}\Theta(-\omega)$$

  • $\begingroup$ Thanks for the correction, I'll edit the question then. $\endgroup$
    – John
    Jan 29 '18 at 14:54
  • $\begingroup$ The sign simply depends on your definition of the Fourier transform; if you take the transform to be the integral of $f(x) e^{i w x}$, you'll get $\mathcal F[1/(1- i x)] = 2 \pi e^w \Theta(-w)$. $\endgroup$
    – Maxim
    Jun 18 '18 at 3:19

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