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 $U(a,b,z) = \frac{G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}[1-a], [] \\ [0,1-b],[]\end{smallmatrix}\right.\right)}{\Gamma\left(a\right)\Gamma\left(1+a-b\right)} = \frac{\frac{1}{2\pi i}\oint \Gamma\left(-s\right) \Gamma\left(a+s\right) \Gamma\left(1+a-b+s\right)z^{-s}ds }{\Gamma\left(a\right)\Gamma\left(1+a-b\right)}$

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)

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

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

Answer 1

There is no issue at all, except that you made a mistake in expanding the Meijer-G function in your first assertion.

Both Abramowitz, Wolfram and Python are correct (and NIST, and Prudnikov et al....)

The wrong bit in your question is:

$$[^*]\; G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}[1-a], [] \\ [0,1-b],[]\end{smallmatrix}\right.\right)= \frac{1}{2\pi i}\oint \Gamma\left(-s\right) \Gamma\left(a+s\right) \Gamma\left(1+a-b+s\right)z^{-s}ds $$

By the very definition of the Meijer-G function (NIST handbook 2010 p.415 16.17.1): $$G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}1-a \\ 0,1-b\end{smallmatrix}\right.\right)=\frac {1}{2\pi i} \int_L \Gamma(-s)\Gamma(a+s)\Gamma(1-b-s) z^s ds$$ So now by using the obvious change of variables $s = -u-a$:

\begin{align} G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}1-a \\ 0,1-b\end{smallmatrix}\right.\right)&=\frac {1}{2\pi i} \int_L \Gamma(u+a)\Gamma(-u)\Gamma(1+a-b+u) z^{-u-a} du \\ &= \frac {1}{2\pi i}z^{-a} \int_L \Gamma(-u)\Gamma(a+u)\Gamma(1+a-b+u) z^{-u} du \end{align} so that $$U(a,b,z) = \frac{G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}1-a \\ 0,1-b\end{smallmatrix}\right.\right)}{\Gamma(a)\Gamma(1+a-b)} = \frac {\int_L \Gamma(-u)\Gamma(a+u)\Gamma(1+a-b+u) z^{-u} du}{2\pi i \Gamma(a)\Gamma(1+a-b)z^a} $$

QED.

Answer 2

In the Wolfram website, it is claimed that $$ U(a, b; z) = \frac{1}{\Gamma(a)\Gamma(a-b+1)}G_{1,2}^{2,1}\left(z\left|\begin{smallmatrix}1-a\\ 0, 1-b\end{smallmatrix}\right.\right). $$ I've checked with Python for a set of values and this looks correct:

from mpmath import *

a = mpc(3, 2)
b = mpc(2, 3)
z = mpc(1, 1)

g = meijerg([[1-a],[]], [[0, 1-b],[]], z) / (gamma(a) * gamma(a-b+1))
u = hyperu(a, b, z)

print(g, u)

gives:

(-0.0673314909897862 - 0.152245341184982j) (-0.0673314909897862 - 0.152245341184982j)
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