Bilateral Laplace transform

Question

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

1. What is the condition for existence of bilateral Laplace transform?
2. How is the condition for existence related to ROC ?
3. What's the bilateral Laplace transform of $\sin \omega t$ ?
4. How is the bilateral Laplace related to Fourier transform and the condition to transform one to another.

Answer 1

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

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

3. $$\mathcal{B}\{\sin a t\}(i s) = \pi i \left( \delta (a-i s)- \delta (a+i s)\right)$$ 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.

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

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