Hello all,
I have a question. Probably it is a very simple question for mathematicians, but I need help to understand it. Suppose F(s) is the Laplace transform of F(t) and F(s) = 1/sqrt(s) exp(-a sqrt(s)). So I want to perform an inverse Laplace transform of F(s) to find F(t). To do this, I define a new variable u = sqrt(s). From that then L^-1{1/u exp(-au)} can be solved easily. However, I am not sure if the result of the inverse Laplace transform is still equal to F(t) or something else! Any comment?
Thanks in advance,
Mehdi
Dear friend Mehdi Rahmati
Yes, you can use a variable substitution to perform the inverse Laplace transform of F(s) = 1/sqrt(s) exp(-a sqrt(s)).
As you mentioned, let u = sqrt(s), then we can write:
F(s) = 1/sqrt(s) exp(-a sqrt(s)) = 1/u exp(-au)
The inverse Laplace transform of 1/u exp(-au) can be found using the following formula:
L^-1{1/u exp(-au)} = 1/t - a L^-1{exp(-at)}
where L^-1{exp(-at)} is the inverse Laplace transform of exp(-at), which is simply δ(t-a), the Dirac delta function shifted to t = a.
Substituting this result back into the formula, we get:
L^-1{1/u exp(-au)} = 1/t - a δ(t-a)
Now, we can substitute back u = sqrt(s) and simplify the expression to obtain the inverse Laplace transform of F(s):
F(t) = L^-1{F(s)} = L^-1{1/sqrt(s) exp(-a sqrt(s))} = L^-1{1/u exp(-au)} = 1/√t - a δ(t-a√)
So the inverse Laplace transform of F(s) is F(t) = 1/√t - a δ(t-a√).
Therefore, the result of the inverse Laplace transform is not equal to F(t), but rather an expression in terms of F(t) and the Dirac delta function.
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