Connection between unilateral and bilateral Laplace transforms

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Jhenrique
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Exist some conection between: $$\int_{0}^{+\infty} f(t) \exp(-st)dt\;\;(1)$$ $$\int_{-\infty}^{0} f(t) \exp(-st)dt\;\;(2)$$ ?

The results, the transformations, are very similar, with some little difference in the signal. So, known the transformation (1), is possible to find the (2)?
 
Jhenrique said:
Exist some conection between: $$\int_{0}^{+\infty} f(t) \exp(-st)dt\;\;(1)$$ $$\int_{-\infty}^{0} f(t) \exp(-st)dt\;\;(2)$$ ?

The results, the transformations, are very similar, with some little difference in the signal. So, known the transformation (1), is possible to find the (2)?
Where did you see the second equation above? The first one is the definition of the Laplace transform of a function f.
 
^There is a bilateral Laplace transform that is the sum of the two unilateral Laplace transforms.
$$\int_{-\infty}^\infty \! \mathrm{f}(t)e^{-s \, t} \, \mathrm{d}x=\int_{-\infty}^0 \! \mathrm{f}(t)e^{-s \, t} \, \mathrm{d}x+\int_{0}^\infty \! \mathrm{f}(t)e^{-s \, t} \, \mathrm{d}x$$

Obviously in general we can say nothing about the two integrals. The function can be totally different for positive and negative values.