What would the laplace inverse of a laplace be?

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SUMMARY

The Laplace inverse of a Laplace transform is defined by the relationship ##\mathcal{L}^{-1}[F(s)] = f(t)##, where ##F(s) = \mathcal{L}[f(t)]##. In the example provided, ##F(s) = L\{t^3 e^{-16t} \cos(3t) \sin^2(t)\}##, the inverse can be computed using the standard definition of the Laplace transform, which is given by the integral ##F = \int_0^\infty e^{-st} f(t) dt##. This establishes a clear method for finding the original function from its Laplace transform.

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Jim wah
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For example:
If F(s) = L{t3e-16tcos(3t)sin2(t)}

What would L-1{F(s)} be?
 
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Jim wah said:
For example:
If F(s) = L{t3e-16tcos(3t)sin2(t)}

What would L-1{F(s)} be?

If ##\mathcal{L}[f(t)] = F(s)##, then ##\mathcal{L}^{-1}[F(s)] = \mathcal{L}^{-1}[\mathcal{L}[f(t)]] = f(t)##
 
Last edited:
[itex]F(s)= L[t3e^{-16t}cos(3t)sin^2(t)][/itex] seem perfectly reasonable to me. A standard definition of the Laplace transform is
[tex]F<s>= \int_0^\infty e^{-st}f(t)dt</s>[/tex]
so that the Laplace transform of a function of t is a function of s.
 
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