1
$\begingroup$

Given,

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

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.

$\endgroup$
0
$\begingroup$

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$$\dfrac{1}{1+it}\Leftarrow\Rightarrow2\pi e^{\omega}\Theta(-\omega)$$

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.