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I have this Laplace transform:

$$\left( w \frac{L}{L+s}+(1-w) \frac{Q}{Q+s}\right)^n \ for \ L>0, Q>0,0<w<1.\ (1)$$

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

Such can also be written as

$$\sum_{k=0}^n\frac{n!}{k!\ (n-k)!}\ w^k\ (1-w)^{n-k}\ \left(\frac{L} {L+s}\right)^k\ \left(\frac{Q}{Q+s}\right)^{n-k}.\ (2) $$

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}]

I've applied inverse Laplace transform of the convolution form on $\left(\frac{L}{L +s }\right)^k\left(\frac{Q}{Q +s }\right)^{n-k}.\ \ (3)$

After performing an inverse Laplace transform on each tern in the equation (3), I then transformed into integral of a convolutional form, $$\frac{L^k Q^{n-k}}{\Gamma (k) \Gamma (n-k)} \int_0^t s^{k-1} e^{-L s} e^{-Q (t-s)} (t-s)^{n-k-1} \, ds\ \ for\ L>0, Q>0, 0\leq k\leq n, n\geq 0.\ \ (4) $$ In code:

(L^k Q^(n-k))/(Gamma[k]Gamma[n-k]) Integrate[E^(-s L) s^(k-1) E^(-(t-s) Q)  (t-s)^(n-k-1),{s,0,t},Assumptions->L>0 &&Q>0&&0<=k<=n&&n>=0]// PowerExpand // Simplify

Such implementation in Mathematica gives me the following answer,

$$L^k Q^{n-k} t^{n-1}\exp (-Qt)\, _ 1\tilde {F} _ 1 (k,n;(Q-L)t)\ \text {if}\ 0<k<n.\ \ (5)$$

Combining with eq (2),

$$\sum_{k=0}^n\frac{n!}{k!\ (n-k)!}\ (L w)^k ((1-w)Q)^{n-k} t^{n-1}\ e^ {-Qt}\, _ 1\tilde {F} _ 1 (k,n;(Q-L)t)\ (6)$$

But here the constraint change i.e. $0<L<Q$, not remain as same as in eq (1).

Is there any way to keep the constraints $L>0, Q>0$ but not $L<Q$?

Please lend a hand.

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  • $\begingroup$ Isn't it that question in other formulas? $\endgroup$
    – user64494
    May 12 at 10:33
  • $\begingroup$ @user64494 However, Mathematica's implementation of that approach is not yet capable of solving the same problem, resulting in a double infinite series too. But here at the expense of modifying the constraints. Do you know of any other options? $\endgroup$ May 12 at 13:05

1 Answer 1

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Too long for a comment. You apply the result of

(L^k Q^(n-k))/(Gamma[k]Gamma[n-k]) Integrate[E^(-s L) s^(k-1) E^(-(t-s) Q)  (t-s)^(n-k-1),{s,0,t},Assumptions->L>0 &&Q>0&&0<=k<=n&&n>=0]// PowerExpand // Simplify

ConditionalExpression[ E^(-Q t) L^k Q^(-k + n) t^(-1 + n) Hypergeometric1F1Regularized[k, n, (-L + Q) t], 0 < k < n]

in

Sum[n!/(k! (n - k)!) w^k (1 - w)^(n - k) *E^(-Q t) L^k Q^(-k + n)
t^(-1 + n) Hypergeometric1F1Regularized[k, n, (-L + Q) t], {k, 0,  n}, Assumptions -> L > 0 && Q > 0]

E^(-Q t) Q^n t^(-1 + n) (1 - w)^n n! DifferenceRoot[...]

Saying nothing about the rearranging the order of summation and integration, for k==n we have

Integrate[ E^(-s L) s^(k - 1) E^(-(t - s) Q) (t - s)^(n - k - 1) /. k -> n, {s, 0, t}, 
Assumptions -> L > 0 && Q > 0 && n >= 0, GenerateConditions -> True]

Integrate::idiv: Integral of (E^(-L s+Q (s-t)) s^(-1+n))/(-s+t) does not converge on {0,t}.

Therefore, your reasoning is not true.

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  • $\begingroup$ Thanks for your input. I'll come back. I need time to understand. $\endgroup$ May 13 at 7:38
  • $\begingroup$ @step-by-step: The pleasure is all mine. Don't hesitate to ask for further explanation in need. $\endgroup$
    – user64494
    May 13 at 8:29

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