Numerical Inversion of Laplace Transforms: 
The Zakian Method
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Acknowledgement :
Rice and Do, Applied Mathematics and Modeling for Chemical Engineers, Wiley, 1995.

Off[General::"spell1"]

Analytical Solution using the built-in function of Mathematica: InverseLaplaceTransform

F[s_] := 1/(s^2 (s + 1)^2)

fexact[t_] := InverseLaplaceTransform[F[s], s, t] // Expand

fexact[t]

plt1 = 
      Plot[fexact[t], {t, 0, 10}, 
        PlotStyle ->  RGBColor[1, 0, 0]]

Numerical Solution using the Zakian method gives the same result as the exact solution

α = {1.283767675 10^1 + 
            I 1.666063445, 
1.222613209 10^1 + 
            I 5.012718792, 
1.09343031 10^1 + 
            I 8.40967312, 
8.77643472 +  
            I 1.19218539 10^1, 
5.22545336 + 
            I 1.57295290 10^1};

K = {-3.69020821 10^4 + 
            I 1.96990426 10^5, 
6.12770252 10^4 - 
            I 9.54086255 10^4, 
-2.89165629 10^4 +  
            I 1.81691853 10^4, 
4.65536114 10^3 -  
            I 1.90152864, 
-1.18741401 10^2 - 
            I 1.41303691 10^2};

TableForm[Transpose[{α, K}]]

t = Table[i, {i, 0.5, 10, 0.5}];

fexact[t]

finv = 
      2/t (∑_(i = 1)^5) Re[
            K[[i]] F[α[[i]]/t]]

TableForm[Transpose[{t, fexact[t], finv}]]

plt2=ListPlot[Transpose[{t, finv}], 
        PlotStyle ->  {PointSize[0.025], 
            RGBColor[0, 0, 1]}]

Show[plt1, plt2]