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    From Stroud's Further Engineering Maths..

    Where L(F(t)) is the Laplace Transform of a function:

    \displaystyle\int^{\infty}_0 {e^{-st}} \, dt = {L(F(t))}

    In determining the transform of any function, you will appreciate that the limits are subsituted for t so that the result will be a function of s:

    \displaystyle\int^{\infty}_0 {e^{-st}} \, dt = {L(F(t))}={f(s)}

    I don't understand the jump from the first equation to the second, how does the Laplace Transform = f(s)?

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    Lee
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    Well your transforming the function from one domain to another. In the first domain the variable is t and in the transformed domain the variable is s, and the laplace transformation maps the function F into the function f.

    Its like if I said  \displaystyle \frac{\partial F(x)}{\partial x} = f(t) , I'm just carrying out an operation/transformation that takes one function into another.

    But the laplace transformation is defined as:

     \displaystyle \mathcal{L}\{F(t)\} = \int_{0}^{\infty} F(t) e^{-s\,t} dt = f(s)

    In the transformed domain things are often easier to deal with (i.e. differential equations just become algebraic equations) so I can form a nice neat equation then do the inverse transformation to get the answer I require in terms of variables of my domain. Also the integral removes any t dependence since it is a definite integral, just as the inverse transformation gives back the t dependence but removes the s dependence.
 
 
 

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