# Fourier Sine Transform

1. Feb 26, 2013

### Arkuski

I am having a hard time understanding how the Fourier sine transform works. I understand that you input a function and you get a function as an output, but I have no idea how certain inputs are even valid. Here is what the book gives for the transform:
$F(\omega )=\frac{2}{\pi}\int_{0}^{\infty}f(t)\displaystyle\sin (\omega t)dt$
For my example, I need to transform an IC in a PDE problem. $u(x,0)=H(1-x)$ where $H$ is the Heaviside function. Basically, $u(x,0)=1$ when $0\le x<1$ and $u(x,0)=0$ when $x\ge 1$. I assume then that the transform works out in the following manner:
$F(\omega )=\frac{2}{\pi}\int_{0}^{1}\displaystyle\sin (\omega t)dt=\frac{2}{\pi \omega}(1-\displaystyle\cos (\omega)$
What would happen in the case that $f(t)=1$? The integral would not converge, yet, mathematica tells me there is an answer.

2. Feb 26, 2013

### jbunniii

If $f(t) = 1$ for all $t \in \mathbb{R}$, you are of course correct. The integral $\int_{0}^{\infty} \sin(\omega t) dt$ does not converge for any $\omega \neq 0$. What "answer" did Mathematica give you?

3. Feb 27, 2013

### Arkuski

I was getting $\sqrt{\frac{2}{\pi}}/\omega$

Last edited: Feb 27, 2013