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.