Connection between unilateral laplace

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SUMMARY

The discussion centers on the relationship between the unilateral Laplace transforms defined by the integrals $$\int_{0}^{+\infty} f(t) \exp(-st)dt$$ and $$\int_{-\infty}^{0} f(t) \exp(-st)dt$$. It is established that the first integral represents the unilateral Laplace transform, while the second integral can be connected through the bilateral Laplace transform, which combines both unilateral transforms. The bilateral Laplace transform is expressed as $$\int_{-\infty}^\infty f(t)e^{-st} dt = \int_{-\infty}^0 f(t)e^{-st} dt + \int_{0}^\infty f(t)e^{-st} dt$$. However, it is noted that without additional information about the function f(t), no definitive conclusions can be drawn regarding the relationship between the two integrals.

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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)?
 
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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.
 

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