How to write Tricomi's confluent hypergeometric function in terms of Meijer-G function

Question

I am calculating a closed form expectation and I encountered the Tricomi's confluent hypergeometric function (aka confluent hypergeometric function of the second kind) given by integral $U\left( a,b,z \right) = \frac{1}{\Gamma\left(a\right)} \int_{0}^{\infty} e^{-zt}t^{a-1}\left(1+t\right)^{b-a-1}$. I need to write this in terms of Meijer-G function $G_{p,q}^{m,n}\left(z\left|\begin{smallmatrix}\mathbf{a}_n, \mathbf{a}_{p-n}\\ \mathbf{b}_m, \mathbf{b}_{q-m}\end{smallmatrix}\right.\right)$.

I used the "MeijerGReduce[HypergeometricU[a, b, z], z] in wolfram alpha and got the solution $U(a,b,z) = \frac{G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}[1-a], [] \\ [0,1-b],[]\end{smallmatrix}\right.\right)}{\Gamma\left(a\right)\Gamma\left(1+a-b\right)} = \frac{\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \Gamma\left(-s\right) \Gamma\left(a+s\right) \Gamma\left(1+a-b+s\right)z^{-s}ds }{\Gamma\left(a\right)\Gamma\left(1+a-b\right)}$

However in the book Handbook of Mathematical Functions: With Formulas, Graphs, and Mathematical Tables it is written in terms of Barnes type integrals as follow (equaion 13.2.10 page 506)

$U(a,b,z) = \frac{\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \Gamma\left(-s\right) \Gamma\left(a+s\right) \Gamma\left(1+a-b+s\right)z^{-s}ds }{\Gamma\left(a\right)\Gamma\left(1+a-b\right)z^a}$

In other words the second one has a $z^a$ term in the denominator! Which one is correct? I appreciate the hints and answers!

P.S. I copied the Wolfram answer as latex and copied it here $\frac{\text{MeijerG}[\{\{1-a\},\{\}\},\{\{0,1-b\},\{\}\},z]}{\Gamma (a) \Gamma (a-b+1)}$. Also the solution as "input text" is Inactive[MeijerG][{{1 - a}, {}}, {{0, 1 - b}, {}}, z]/( Gamma[a] Gamma[1 + a - b]). Furthermore added the part from the book too!

This is also the figure that shows the solution using Wolfram Mathematica

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I asked this same equation here at the math.stackexchange and the correct answer with proof of integral is over there. I often ask my math questions there, however sometimes the harder question that are related to higher math, I ask them here too. Thank you!

Answer 1

They are both correct.

Numerical check of the expression: $U(a,b,z) = \frac{G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}[1-a], [] \\ [0,1-b],[]\end{smallmatrix}\right.\right)}{\Gamma\left(a\right)\Gamma\left(1+a-b\right)} \text{with some condition}$

Clear["Global`*"];
HypergeometricU[a, b, z] /. {a -> 2.33, b -> 3.44, z -> 4.55}
(* 0.0305946 *)



HypergeometricU[a, b, z] // MeijerGReduce[#, z] &   
(* Inactive[MeijerG][{{1 - a}, {}}, {{0, 1 - b}, {}}, z]/(Gamma[a] Gamma[1 + a - b]) *)

% /. {a -> 2.33, b -> 3.44, z -> 4.55} // Activate // N
(* 0.0305946 *)

 

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Check of eqn(13.2.10) in page 506 from Handbook of Mathematical Functions: With Formulas, Graphs, and Mathematical Tables:

13.2.10

$$ \Gamma(a)\Gamma(1+a-b)z^aU(a,b,z) \\ = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(-s)\Gamma(a+s)\Gamma(1+a-b+s)z^{-s}ds$$ for $|\arg z| \lt \frac{3\pi}{2}$, $a\neq0,-1,-2,\ldots$, $b-a \neq 1,2,3,\ldots$. The contour must separate the poles of $\Gamma(-s)$ from those of $\Gamma(a+s)$ and $\Gamma(1+a-b+s)$.

Clear["Global`*"];

RHS = InverseMellinTransform[
    Gamma[-s]*Gamma[a + s]*Gamma[1 + a - b + s], s, z, 
    Assumptions -> {Abs[Arg[z]] < 3*Pi/2,  
       !Element[a, NonPositiveIntegers],  
       !Element[b - a, PositiveIntegers]}] // FullSimplify;
LHS = Gamma[a]*Gamma[1 + a - b]*z^a*HypergeometricU[a, b, z];

{RHS, LHS}

{ConditionalExpression[
  z^a Gamma[a] Gamma[1 + a - b] HypergeometricU[a, b, z], 
  Max[-Re[a], -1 - Re[a] + Re[b]] < Re[s] < 0], 
 z^a Gamma[a] Gamma[1 + a - b] HypergeometricU[a, b, z]}