# Fourier Transforms - Tidying Up After Integration

1. May 5, 2012

### Saz1

ω1. The problem statement, all variables and given/known data

The function f(x) is defined by:

f(x) = e^2ax when x ≤ 0
0 when x > 0

Show that, for a > 0, its fourier transform may be written:

fourier transform = 1 / (2a - iω)

2. Relevant equations

fourier transform = ∫f(x)e^iωt dx

(the integral is taken over minus infinity to infinity).

3. The attempt at a solution

I think I have done the 'hard' part correct:

fourier transform = ∫e^(2ax) × e^iωt

= ∫e^(2a-iω)x

= e^(2a-iω)x / (2a-iω)x

but how would I go about getting it in the form required? Im guessing it related to the fact that the integral is taken over minus infinity to infinity - as x tends to infinity the fourier transform tends to zero.. as x tends to minus infinity.. ?? What happens to the exponent?

Also, as a side note, is what I have done so far correct? Thanks alot :)

2. May 5, 2012

### vela

Staff Emeritus
You need to get in the habit of using parentheses. Also, you're being sloppy with x and t. Decide on one or the other.

Since the function is defined piecewise, break the integral up accordingly:
$$\int_{-\infty}^{\infty} f(t)e^{-i\omega t}\,dt = \int_{-\infty}^{0} f(t)e^{-i\omega t}\,dt + \int_{0}^{\infty} f(t)e^{-i\omega t}\,dt$$ Now what do you get?

3. May 5, 2012

### Ray Vickson

For x < 0, is your function $e^2 ax, \, e^{2a}x, \text{ or } e^{2ax}\:?$ If you mean the third one you either need to use brackets, like this e^(2ax) or use the "superscript" button, like this e2ax. That button is found on the pallette at the top of the message area, and looks like X2.

RGV

4. May 7, 2012

### Saz1

Sorry, first time I've posted here and I'm just generally rubbish with computers. There's no 't' value - everything is in terms of x. So the second integral would equal zero as defined by the initial function, and the first integral taken over minus infinity to zero would become

1/(2a-iω) × e(2a-iω)x

then when you put x=0 in, e^0 = 1 and hence the integral would be in the form I require. Thanks alot for your help.