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.

Check the book "Advanced Engineering Mathematics" by Erwin Kreyszig