Bilateral Laplace transform My knowledge of Bilateral Laplace transform is less. Here are the few questions I need answer.


*

*What is the condition for existence of bilateral Laplace transform?

*How is the condition for existence related to ROC ?

*What's the bilateral Laplace transform of $\sin \omega t$ ?

*How is the bilateral Laplace related to Fourier transform and the condition to transform one to another.

 A: *

*A sufficient condition that $f(t)$ has bilateral Laplacian transform $\mathcal{B}\{f(t)\}(s)$ if $e^{-st}f(t)$ absolute integrable, i.e. 
$$\int_{-\infty}^\infty |e^{-s t}f(t)|=\int_{-\infty}^\infty e^{-Re(s)t}|f(t)|<\infty$$

*Range of convergence of bilateral Laplacian transform is defined 
$$ \mathrm{ROC}=\left\{s\Bigg| \int_{-\infty}^\infty e^{-Re(s)t}|f(t)|<\infty\right\}$$

*$$\mathcal{B}\{\sin a t\}(i s) = \pi i \left( \delta (a-i s)- \delta
    (a+i s)\right)$$ where $\delta(x)$ is a Dirac Delta Function
In detail:
I have found in some table that  $$\int_{-\infty}^\infty
e^{-ixy}f(ax)\sin bx dx = \frac{1}{2ai}\left( \int_{-\infty}^\infty
e^{-ix{\frac{y-b}a}}f(x) dx- \int_{-\infty}^\infty
e^{-ix{\frac{y+b}a}}f(x) dx \right)$$  
In your case $f(x)=1$, $a=1$, $b=\omega$, so $$ \mathcal{B}\{\sin
at\}({is})= \int_{-\infty}^\infty e^{-its}\sin at dt =
\frac{1}{2i}\left( \int_{-\infty}^\infty e^{-it{{(s-a)}}} dt-
\int_{-\infty}^\infty e^{-it{{(s+a)}}} dt \right)$$ which is the
the answer above is obtained since
$$\delta(x-a)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-it{{(x-a)}}}
dx,$$ see  here  ,here and here.

*The continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform with imaginary argument $s = iω$ or $s = 2πfi$:$$
  \hat{f}(\omega) = \mathcal{F}\{f(t)\} 
              = \mathcal{B}\{f(t)\}|_{s = i\omega}  =  F(s)|_{s = i \omega} 
              = \int_{-\infty}^{\infty} e^{-i \omega t} f(t)\,dt. \\
$$

You might be interesting in a "Advanced Mathematical Analysis" book by Richard Beals.
