# Fourier Transform of Heaviside function

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1. Mar 24, 2016

### EdisT

1. The problem statement, all variables and given/known data
Find the Fourier transform of $$H(x-a)e^{-bx},$$ where H(x) is the Heaviside function.

2. Relevant equations
$$\mathcal{F}[f(t)]=\frac{1}{2 \pi} \int_{- \infty}^{\infty} f(t) \cdot e^{-i \omega t} dt$$
Convolution theory equations that might be relevant:
$$\mathcal{F}[f(t) \cdot g(t)] = \mathcal{F}[f(t)] * \mathcal{F}[g(t)]$$
$$\mathcal{F}[e^{\alpha t} \cdot f(t)] = \mathcal{F}[f(t+\alpha t)]$$
Derivative of the Heaviside function:
$$H'(t) = \delta(t)$$
where $\delta(t)$ is the Dirac Delta function.

3. The attempt at a solution
Using the first relevant equation, and assuming the Heaviside function simply changed the boundary conditions from a to infinity:
$$\mathcal{F}[H(x-a)e^{-bx}]=\frac{1}{2 \pi} \int_{a}^{\infty} e^{-bx} \cdot e^{-i \omega t} dt$$
$$= - \frac{1}{2 \pi} \left[\frac{i e^{-bx-i \omega t}}{\omega} \right]_{a}^{\infty} = - \frac{i e^{-bx-i \omega a}}{\sqrt{2\pi} \omega}$$

This was not right so I tried to take the integral of the Heaviside function via integration by parts:
$$\mathcal{F}[f(t)]=\frac{1}{2 \pi} \int_{- \infty}^{\infty}H(x-a)e^{-bx-i \omega t} dt$$
$$=\frac{1}{2 \pi} \left[\frac{-H(x-a)ie^{-bx-i \omega t}}{\omega} \right]_{-\infty}^{\infty} - \int_{\infty}^{\infty} \delta(x-a)e^{-bx-i \omega t}$$
At this point I could go on with the integration however I'm unsure if it'll lead to the right answer or if what I'm doing makes sense. I have considered using the last equation in part of the convolution theory section, to get an integral of perhaps $H(x-a-i \omega t)$ but that doesn't seem right either.. I am unsure how to take the Fourier transformation of the Heaviside function, any help will be gladly appreciated.

2. Mar 24, 2016

### Orodruin

Staff Emeritus
You are mixing variables. Either use x or t, for this problem it is quite obvious that x is the variable and you need to set up your Fourier transform accordingly. (If it was in t, then your function would be a constant and the Heaviside would not change the integration domain.)

3. Mar 24, 2016

### EdisT

Would it then just be the integral from a to infinity of $H(x-a)e^{-bx-i \omega x} dx$ or the integral from a to infinity of $e^{-bx-i \omega x}$ ?

4. Mar 24, 2016

### Ray Vickson

Aren't these exactly the same thing?