Tried for Inverse Laplace transform (ILT) for the following:

L[s] = (L /(L + s) w + Q /(Q + s) (1 - w))^n 

$L[s]$ can also be written as

$$L[s]=\sum_{k=0}^n\frac{n!}{k!\ (n-k)!}\ w^k\ (1-w)^{n-k}\ \left(\frac{L} {L+s}\right)^k\ \left(\frac{Q}{Q+s}\right)^{n-k}\ (1) $$

In code:

Sum[n!/(k! (n - k)!) w^k (1 - w)^(n - k) (L/(L + s))^k (Q/(Q + s))^(n - k), {k, 0, n}]

The ILT of auxiliary variable $s$ is referred from Tables of Integral Transforms, Vol. 1, and the complete ILT is given as $$g(t)=\sum_{k=0}^n\frac{n!}{k!\ (n-k)!}\ w^k\ (1-w)^{n-k}\ L^k\ Q^{n-k}\ t^{n-1}\ \phi_2(k, n - k, n -L\ t, -Q\ t)\ (2)$$

where $\phi_2(*)$ is hyper-geometric series with $$\phi_2(A,B,C, x,y)=\sum_{u=0}^\infty \sum_{v=0}^\infty \frac{(A)_u(B)_v x^u y^v}{(C)_{u+v} u! v!},\quad where (*)_a = Pochammer[*,a]\quad (3)$$

The equivalent code of $(3)$ is,

Sum[(Pochhammer[A,u] Pochhammer[B,v] (x^u y^v))/(Pochhammer[C,u+v] u! v!), {u,0, ∞}, {v,0, ∞}]

If look into "Tables of Integral Transforms, Vol. 1, pg-238 (PDF pg-253; form 9)", then given Laplace transform is

$$\Gamma(\gamma)\ p^{-\gamma}\left(1-\left(\frac{\lambda_1}{p}\right)^{-\beta_1}\right) {...}\left(1-\left(\frac{\lambda_n}{p}\right)^{-\beta_n}\right), Re\ \gamma>0\quad (4)$$

inverted as

$$t^{\gamma-1}\ \phi_2(\beta_1,{...},\beta_n;\gamma;\lambda_1\ t,{...},\lambda_n\ t)\quad (5)$$

Following ways tried,

  1. If no replacement mistake, then $(4)$ taken in terms of $(1)$ variables as (ignoring terms without $s$), $\lambda_1\to L, \lambda_2\to Q, p\to s, \gamma \to n, \beta_1 \to k$ and $\beta_2 \to n-k$.

To check such,

InverseLaplaceTransform[Gamma[n]/s^n (1-L/s)^k (1-Q/s)^(n-k),s,t]

We got, Gamma[n] InverseLaplaceTransform[1/s^n (1-L/s)^k (1-Q/s)^(n-k),s,t]

Wondering, is Mathematica didn't recognized the known form or form to be modified for evaluation ?

  1. Then, computing with direct integrate for

    H = (L /(L + s))^k (Q/(Q + s))^(n - k);
    1/(2 π j) Integrate[H Exp[s t], {s, c - j ∞, c + j ∞}, Assumptions -> c > 2 && t > 0]
    (*output same as input*) 

    What could be the reason?

  2. Furthermore, then convolution theorem applied,

    list = {(L/(L + s))^k, (Q/(Q + s))^(n - k)};
    intg = Times@@(Map[InverseLaplaceTransform[#, s, t] &,list]//{#[[1]]/.t -> s,#[[2]]/.t -> (t-s)}&);
    Integrate[intg, {s,0,t}] // PowerExpand
    (*E^(-Q t) L^k Q^(-k+n) t^(-1+n) Hypergeometric1F1Regularized[k,n,(-L+Q) t]*)

It gave out something, but not near to what $(2)$ contained.

  1. Further searched that,AppellF1[A;B,C;c;x,y] form could be related to $(3)$. But not exactly, as the equivalent form of $\phi_2(k, n - k, n -L\ t, -Q\ t)$ is not clearly found in Wolfram documents.

All trials does not work out.

So kindly help on, how we can evaluate $(1)$ to get $(2)$, by InverseLaplaceTransform[.] or Integrate[.]?

  • $\begingroup$ Mathematica didn't recognized this $\phi_2$ function ,because is not implemented. Why do you need exactly this function? $\phi_2$ is Horns function see: mathworld.wolfram.com/HornFunction.html $\endgroup$ Jul 1, 2018 at 12:36
  • $\begingroup$ @MariuszIwaniuk thanks. Two of the articles, mentioned $(1)$ solved to $(2)$. So I am solving it for another similar modelling. But could not get through either mentioned ways. Terrible! if Mathematica don't have this. Any other alternatives? $\endgroup$ Jul 2, 2018 at 5:52
  • 1
    $\begingroup$ At a given value 'n' can be found a solution to the ILT. About alternatives,well I don't know.Maple also has no this function. $\endgroup$ Jul 2, 2018 at 6:20
  • $\begingroup$ @MariuszIwaniuk. I will try it. Wanted to learn more, can you please suggest any, less mathematical, article on Horns function ? :) $\endgroup$ Jul 3, 2018 at 12:44
  • $\begingroup$ See REFERENCES at the end of the page: mathworld.wolfram.com/HornFunction.html $\endgroup$ Jul 3, 2018 at 13:00


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