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In my calculations, we must reverse the Laplace transform of the hypergeometric confluent function. In the (integrals and series, inverse Laplace transform)A.P. Prudnikov's book , the inverse Laplace function is subtracted from the hypergeometric function as follows.
$$L^{-1}[frac{Γ(p+a)}{Γ(p+b)}1F1(p+a,p+b,w)=frac{(1-e^{-x})^{b-a-1}}{Γ(b-a)}exp(we^{-x}-ax)]$$
The function that I have to get from Laplace's reverse is as follows.
$$L^{-1}[frac{Γ(p+a)}{Γ(2p+b)}1F1(p+a,2p+b,w)$$
To use the formula in mentioned book, how should I convert this function to the hypergeometric function shown in this book?
In fact, I want to get the inverse Laplace transform from the following statement
$$L^{-1}[1F1(as+b,2as+1,x)]$$
we have
$$sum_{n=0}^inftyfrac{x^{n}}{n!}L^{-1}(frac{(as+b)_{n}}{(2as+1)_{n}})=sum_{n=0}^inftyfrac{x^{n}}{n!}L^{-1}(frac{(as+b)(as+b+1)(as+b+2)...(as+b+n-1)}{(2as+1)(2as+2)(2as+3)...(2as+n)})$$ We have the following poles$s_{1}=frac{-1}{2a},s_{2}=frac{-1}{a},...,s_{n}=frac{-n}{2a}$
$$L^{-1}(frac{(as+b)(as+b+1)(as+b+2)...(as+b+n-1)}{(2as+1)(2as+2)(2as+3)...(2as+n)})=frac{1}{2a}(frac{(frac{-1}{2}+b)(frac{1}{2}+b)...(-frac{3}{2}+b+n)}{1*2*...(n-1)}e^{-frac{t}{2a}}+frac{(b-1)(b)...(-2+b+n)}{(-1)*(1)*(2)*(3)...(n-2)}e^{-frac{2t}{2a}}+frac{(frac{-3}{2}+b)(frac{-1}{2}+b)...(-frac{3}{2}+b+n-1)}{(-2)*(-1)*(1)*(2)*...(n-3)}e^{-frac{3t}{2a}}+...+frac{(frac{-n}{2}+b)(frac{-n}{2}+b+1)...(-frac{n}{2}+b-1)}{(-n+1)*(-n+2)*...(-2)*(-1)}e^{-frac{nt}{2a}})$$
$$L^{-1}[1F1(as+b,2as+1,x)]=sum_{n=0}^inftyfrac{x^{n}}{n!}[frac{1}{2a}(frac{(frac{-1}{2}+b)(frac{1}{2}+b)...(-frac{3}{2}+b+n)}{1*2*...(n-1)}e^{-frac{t}{2a}}+frac{(b-1)(b)...(-2+b+n)}{(-1)*(1)*(2)*(3)...(n-2)}e^{-frac{2t}{2a}}+frac{(frac{-3}{2}+b)(frac{-1}{2}+b)...(-frac{3}{2}+b+n-1)}{(-2)*(-1)*(1)*(2)*...(n-3)}e^{-frac{3t}{2a}}+...+frac{(frac{-n}{2}+b)(frac{-n}{2}+b+1)...(-frac{n}{2}+b-1)}{(-n+1)*(-n+2)*...(-2)*(-1)}e^{-frac{nt}{2a}})]=sum_{n=0}^inftyfrac{x^{n}}{n!}[frac{1}{2a}frac{(-frac{1}{2}+b)_{n}}{(n-1)!}e^{-frac{t}{2a}}-frac{(-1+b)_{n}}{(n-2)!}e^{-frac{2t}{2a}}+frac{1}{2}frac{(-frac{3}{2}+b)_{n}}{(n-3)!}e^{-frac{3t}{2a}}+...+frac{(-frac{n}{2}+b)_{n}}{(-1)^{n-1}???}e^{-frac{nt}{2a}}]$$Is my solution correct? Is there a closed solution?What should I place on the question mark?
I also used the other method as follows:
$$L^{-1}[1F1(as+b,2as+1,x)]=sum_{n=0}^inftyfrac{x^{n}}{n!}(L^{-1}{frac{Γ(as+b+n)Γ(2as+1)}{Γ(as+b)Γ(2as+1+n)}})$$there is two poles $s_{1}=frac{-k-b-n}{a}$ and $s_{2}=frac{-k-1}{2a}$ thus, we have
$$sum_{n=0}^inftyfrac{x^{n}}{n!}(sum_{k=0}^inftyfrac{(-1)^{k}Γ(-2k-2b-2n+1)}{k!Γ(-k-n)Γ(-2k-2b-n+1)}e^{(frac{-k-b-n}{a})t}+frac{(-1)^{k}Γ(frac{-k-1}{2}+b+n)}{k!Γ(frac{-k-1}{2}+b)Γ(-k+n)}e^{(frac{-k-1}{2a})t})$$
What do I do for $Γ(-k+n)$ and $Γ(-k-n)$ in the denominator?gamma function for the negative integer numbers is diverge.
 

Let $a > 0, ,b > 0, ,0 < x < 1$. The Bromwich integral of $F(s) = {_1hspace{-2px}F_1}(a s + b; 2 a s + 1; x) - e^{x/2}$ exists and can be computed by applying the residue theorem. Let $j in mathbb N^0, ,l = (j + 1)/2$, then
$$c_{k, j} = operatorname*{Res}_{s = -l/a}
frac {(a s + b)_k} {(2 a s + 1)_k} frac {x^k} {k!} e^{t s} =
frac {(-1)^j (b - l)_k ,x^k e^{-l t/a}} {2 a k! j! (k - j - 1)!}, \
mathcal L^{-1}[F] =
sum_{k = 0}^infty sum_{j = 0}^{k - 1} c_{k, j} =
sum_{j = 0}^infty sum_{k = j + 1}^infty c_{k, j} = \
frac 1 {2 a} sum_{j = 0}^infty
frac {Gamma(b + l)} {Gamma(b - l)}
frac {(-1)^j x^{j + 1}} {j! (j + 1)!}
{_1hspace{-2px}F_1}(b + l; j + 2; x) e^{-l t/a}, \
mathcal L^{-1}_{s to t} [F(s) + e^{x/2}] =
mathcal L^{-1}_{s to t} [F(s)] + e^{x/2} delta(t).$$