Fourier Transform of Heaviside function

[EdisT]

Homework Statement

Find the Fourier transform of H(x-a)e^{-bx}, where H(x) is the Heaviside function.

Homework 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.

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.

 

[Orodruin]

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.)

 

[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} ?

 

[Ray Vickson]

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} ?

Aren't these exactly the same thing?