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


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

  • 2
    $\begingroup$ Welcome to Mathematica StackExchange! Please edit the question and copy-paste your Mathematica code, so that others can reproduce your results. $\endgroup$
    – Domen
    Commented Jun 8 at 16:13
  • $\begingroup$ Thanks for the comment, I edited the question! $\endgroup$ Commented Jun 8 at 16:18

1 Answer 1



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


The integral definition used by Mathematica is

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

enter image description here

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, 

enter image description here

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

enter image description here

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

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

(* True *)
  • $\begingroup$ Thank you for this answer, it is true that it uses $e^{-st}$ in place of $e^{st}$, cause that's the definition of the Laplace transform. I have no problem with that part and the only difference that flipping the sign makes, is that in formulas $s$ turns into $-s$. However, I am using Laplace transform to calculate the moment generating function (MGF) and that's why I used $e^{st}$ but my problem is not with $e^{st}$ or $e^{-st}$. My problem is the term $\cos(a\pi)$, instead of $\cos(a\pi)$ we must have $(-1)^{-a}$. I appreciate your answer and effort sir, however, it does not address my issue! $\endgroup$ Commented Jun 9 at 11:32

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