Can the product of an exponential and a hypergeometric function be expressed as a Meijer-G function?

Let $0<\alpha< 1$. Can the function

$\qquad e^{-\alpha x}U(a,b,x),$

where $U(a,b,z)$ is the hypergeometric U function be expressed as a Meijer-G function?

The closest I found was here which corresponds to $\alpha=1$.

• This seems more a math problem than a Mathematica problem. Dec 8 '17 at 18:22

MeijerGReduce[E^(-z α) HypergeometricU[a, b, z], z]

$$\frac{\text{MeijerG}\left(\{\{\},\{\}\},\left\{\left\{0,\frac{1}{2}\right\},\{\}\right\},\frac{\alpha z}{2},\frac{1}{2}\right) \text{MeijerG}(\{\{1-a\},\{\}\},\{\{0,1-b\},\{\}\},z)}{\sqrt{\pi } \Gamma (a) \Gamma (a-b+1)}$$
I have constrained $0< \alpha< 1$, and I also got the same result.
• I was looking to see if it can be written as a single Meijer-G function. I have an integral of the form $\int_{0}^{\infty}e^{it/x}e^{-\alpha x} U(a,b,x)\mathrm{d}x$. I am looking to hopefully write the integrand as the product of two Meijer-G functions to help solve it. Dec 8 '17 at 18:03
• Mellin Transform gives a sum of two Hypergeometric2F1Regularized functions. You could try to express them as Meijerfunctions, and see what your integral does. Dec 8 '17 at 18:14