'(a) Show that the inverse Laplace transform of F(s) = log is given by fi(x) =-x: Do the same for Fz(s) = log(s + 1) to obtain f2(x) = ~e-x/x_'
Added by Cassandra I.
Step 1
In our case, F(s) = log(s), so we have: L^-1{log(s)} = -t For F(s) = log(s+1), we can use a similar approach: L^-1{F(s)} = L^-1{log(s+1)} = e^(-x)/x To see why this is true, we can use the property that: L{f(t-a)u(t-a)} = e^(-as) F(s) where u(t) is the unit Show more…
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