What is the Fourier Sine Transform of 1?

[Hoplite]

Homework Statement

I'm looking to determine the Fourier sine transfom of 1.

Homework Equations

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

\frac{2}{\pi \omega}

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?

 

[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.

 

[Hoplite]

Excellent. Thanks, LCKrutz.

 

[vela]

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

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.

 

[Hoplite]

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.

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

Cheers.