I am no expert on Laplace transforms, I just tried out the built-in commands of Mathematica:

f[t_] = InverseLaplaceTransform[(12  (1 - E^(168  s) + 
       168  E^(180  s)  s))/(-1 + E^(12  s) + E^(168  s) - 
     E^(180  s) + 2016  E^(204  s)  s^2), s, t]

gives no closed-form output. I tried numerical values:

In[1986]:= {f[1.], f[10.], f[1000.], 16/19.}

Out[1986]= {1.27295*10^-111, -1.82483*10^-10, 0.8421, 0.842105}

which partially agree with what you know.

I had copied and pasted from your second line. Now it's better:

f[t_] = InverseLaplaceTransform[(12  E^(24 s) 
      (1 - E^(168   s) + 168   E^(180   s)   s))/
    (-1 + E^(12   s) + E^(168   s) -
      E^(180   s) + 2016   E^(204   s)   s^2), s, t];
{f[1.], f[10.], f[1000.], 16/19.}

{1., 0.999996, 0.842102, 0.842105}

I have used GWR for many years, with very good results. This is the illusttration how to use it:

(*http://library.wolfram.com/infocenter/MathSource/4738/*)GWR[F_,t_,M_: 32,precin_: 0]:=Module[{M1,G0,Gm,Gp,best,expr,τ=Log[2]/t,Fi,broken,prec},If[precin<=0,prec=21 M/10,prec=precin];
If[prec<=$MachinePrecision,prec=$MachinePrecision];
broken=False;
If[Precision[τ]<prec,τ=SetPrecision[τ,prec]];
Do[Fi[i]=N[F[i τ],prec],{i,1,2 M}];
M1=M;
Do[G0[n-1]=τ (2 n)!/(n! (n-1)!) Sum[Binomial[n,i] (-1)^i Fi[n+i],{i,0,n}];
If[Not[NumberQ[G0[n-1]]],M1=n-1;G0[n-1]=.;Break[]];,{n,1,M}];
Do[Gm[n]=0,{n,0,M1}];
best=G0[M1-1];
Do[Do[expr=G0[n+1]-G0[n];
If[Or[Not[NumberQ[expr]],expr==0],broken=True;Break[]];
expr=Gm[n+1]+(k+1)/expr;
Gp[n]=expr;
If[OddQ[k],If[n==M1-2-k,best=expr]];,{n,M1-2-k,0,-1}];
If[broken,Break[]];
Do[Gm[n]=G0[n];G0[n]=Gp[n],{n,0,M1-k}];,{k,0,M1-2}];
best]
SetAttributes[GWR,{Listable,NHoldAll}]

F[s_]:=12 (1-Exp[168 s]+168 Exp[180 s] s)/(-1+Exp[12 s]+Exp[168 s]-Exp[180 s]+2016 Exp[204 s] s^2);

t={10^-1,1,10,100,1000,10000,100000};

N[GWR[F,t,60],20]
N[GWR[F,t,80],20]
N[GWR[F,t,100],20]
N[GWR[F,t,120],20]
N[GWR[F,t,140],20]

You should increase the number of digits (the third calling parameter of GWR) until you get a satisfactory accuracy of the results. I notice that your original formula has one extra parenthesis, so that I am not sure if after removing it the formula for F is correct. But if it is, then I see that your transform may require a rather high value of the third parameter. Of course, instead of using a list of time values (t) you can just put a single value. Leslaw

See attached file.

Attachments:

InverseLaplace ver3.nb

Thank you so much Mariuz! I think you and Gianluca ignored $e^{24 s}$ in the numerator. Could you please make sure you have used the function given in OP, and see what will happen next? It is also given here.

Which method is the best (you already used Durbin)?

Mariusz Thanks a ton! I am still struggling to run the code in a notebook in ww.wolframcloud.com, but I have been unsuccessful. The following can be compiled (while I could not plot f[t_])

f[t_] = InverseLaplaceTransform[(12 E^(24 s) (1 - E^(168 s) + 168 E^(180 s) s))/ (-1 + E^(12 s) + E^(168 s) - E^(180 s) + 2016 E^(204 s) s^2), s, t].

However, I am getting the error "No Wolfram Language translation found" with your code (I guess I should work in another environment and sorry if I do not have sufficient experience on working with Wolfram environments and Mathematica). To make sure that the output is correct, could you please provide a figure of $p(t)=1-q(t)$ over t= 1 to 1500 (at time 1, 2, 3, ...., 400, and 450, 500, ..., 1500) using "Weeks" method (which seems to be one of most stable methods)? I guess the function have significant fluctuations before 400 ( $p(t)=0$ for $0<t<12$ and tends to 3/19 in at infinity).

I have verified the GWR package many times. You should run it yourself, otherwise you will not get any experience with it. One thing that worries me, regarding your transform, is that due to the presence of exponential terms it can be rather sensitive to variations of s. There is also a subtraction of exponential terms, which for large s (corresponding to small t) can possibly cause a considerable loss of significant digits. It may be necessary to do some rearrangements of your transform, if you need accurate results, but I have no concrete suggestions how to rearrange. Leslaw

Thanks! I just updated my answer here. You may check it.