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 $\alpha=1$.
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Sign up to join this communityLet $0<\alpha< 1$. 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 $\alpha=1$.
Are you asking for this?
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.
Hypergeometric2F1Regularized
functions. You could try to express them as Meijer
functions, and see what your integral does.
$\endgroup$
Dec 8, 2017 at 18:14