# Fourier transform that gives Heaviside step function

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.

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$$\dfrac{1}{1+it}\Leftarrow\Rightarrow2\pi e^{\omega}\Theta(-\omega)$$ • Thanks for the correction, I'll edit the question then. – John Jan 29 '18 at 14:54 • 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)\$. Jun 18 '18 at 3:19