Laplace transform of special function

Question

The Confluent hypergeometric function of first kind (aka Kummer's function) is defined as

$${\mathbf{M}}\left(a,b,z\right)=\frac{1}{\Gamma\left(a\right)\Gamma\left(b-a% \right)}\int_{0}^{1}e^{zt}t^{a-1}(1-t)^{b-a-1}\,\mathrm{d}t$$

I know from NIST-DLMF integral results (equation 13.10.4) that

$$\int_{0}^{\infty}e^{-zt}t^{b-1}{\mathbf{M}}\left(a,b,t\right)\,\mathrm{d}t=z^{% -b}\left(1-\frac{1}{z}\right)^{-a}, \Re\{b\} >0,\Re\{z\} > 1$$

and I want to calculate the integral

$$\int_{0}^{\infty}e^{st}e^{-kt}t^{b-1}{\mathbf{M}}\left(a,b,t\right)\,\mathrm{d}t$$

by comparing and substitution, I calculated the results as

$$\int_{0}^{\infty}e^{st}e^{-kt}t^{b-1}{\mathbf{M}}\left(a,b,t\right)\,\mathrm{d}t = (s-k+1)^{-a} (k-s)^{a-b} , \Re\{b\}>0, \Re\{k-s\}>1$$

However when I plug the integral into Wolfram Mathematica 13.2, I get the following result

$$\int_{0}^{\infty}e^{st}e^{-kt}t^{b-1}{\mathbf{M}}\left(a,b,t\right)\,\mathrm{d}t = \cos (\pi a) (-k+s+1)^{-a} (k-s)^{a-b}\text{ if }, \Re\{b\}>0$$

How is this possible?!? What is going on? Did I made a mistake?

The command I used in Mathematica is as follow

Integrate[Exp[s*t]*Exp[-k*t]*t^(b - 1)*Hypergeometric1F1[a,b,t], {t,0,Infinity}]

There is also link to mathematics stack exchange for this same question cause I thought maybe my integration is wrong.

Thanks in advance

Answer 1

$Version

(* "14.0.0 for Mac OS X ARM (64-bit) (December 13, 2023)" *)

Clear["Global`*"]

The integral definition used by Mathematica is

(def = Entity["MathematicalFunction", "Hypergeometric1F1"][
     "IntegralRepresentations"][[1]]) // TraditionalForm

Assuming[Re[b] > Re[a] > 0, def[a, b, z] // Activate // FullSimplify]

(* True *)

int = Integrate[
  E^(s*t) E^(-k*t) t^(b - 1) Hypergeometric1F1[a, b, t], {t, 0, 
   Infinity}]

lt = LaplaceTransform[E^(-k*t) t^(b - 1) Hypergeometric1F1[a, b, t], t, s,
  GenerateConditions -> True]

Since LaplaceTransform uses E^(-s*t) in place of your E^(s*t)

int == (lt /. s -> -s) // Normal

(* True *)