# Fourier sine transform of 1

1. Jan 11, 2010

### Hoplite

1. The problem statement, all variables and given/known data
I'm looking to determine the Fourier sine transfom of 1.

2. Relevant equations
One this site http://mechse.illinois.edu/research/dstn/teaching_files2/fouriertransforms.pdf [Broken] (page 2) it gives the sine transform as

$$\frac{2}{\pi \omega}$$

3. The attempt at a solution
However, since the Fourier sine fransform of 1 is defined via,

$$\frac{2}{\pi} \int_0^\infty \sin (\omega x) dx ,$$

I figure that its value should be,

$$\frac{2}{\pi \omega} -\lim_{L\rightarrow \infty } \frac{2}{\pi \omega} \cos (r L) .$$

It seems like they've just thrown the cosine term away, but is this legal? If so why?

Last edited by a moderator: May 4, 2017
2. Jan 11, 2010

### LCKurtz

The usual condition for any fourier transform is

$$\int_{-\infty}^\infty |f(x)|\ dx < \infty$$

which f(x) = 1 doesn't satisfy. The sine transform doesn't exist, and the integral for it diverges as you have observed.

3. Jan 11, 2010

### Hoplite

Excellent. Thanks, LCKrutz.

4. Jan 11, 2010

### vela

Staff Emeritus
The straightforward integral diverges, so what they probably did was throw in an integrating factor $e^{-\lambda x}$ to make the integral converge, and then take the limit as $\lambda\rightarrow0^+$. Try that and see what you get.

5. Jan 11, 2010

### Hoplite

That's a good trick. I'll have to remember that one.

Cheers.