Understanding the Fourier Sine Transform: Valid Inputs and Applications

[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.

 

[jbunniii]

What would happen in the case that f(t)=1? The integral would not converge, yet, mathematica tells me there is an answer.

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?

 

[Arkuski]

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