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 $\alpha=1$.
Can the product of an exponential and a hypergeometric function be expressed as a Meijer-G function?
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1$\begingroup$ This seems more a math problem than a Mathematica problem. $\endgroup$ – m_goldberg Dec 8 '17 at 18:22
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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.
answered Dec 8 '17 at 17:56
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$\begingroup$ 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. $\endgroup$ – Aaron Hendrickson Dec 8 '17 at 18:03
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$\begingroup$ Thought the Mellin transform approach could be used as in this question: mathematica.stackexchange.com/questions/153852/… $\endgroup$ – Aaron Hendrickson Dec 8 '17 at 18:06
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$\begingroup$ Mellin Transform gives a sum of two
Hypergeometric2F1Regularizedfunctions. You could try to express them asMeijerfunctions, and see what your integral does. $\endgroup$ – José Antonio Díaz Navas Dec 8 '17 at 18:14 -
$\begingroup$ I suggest you move your question to mathematics.stackexchange.com. Once you have a clear path to do it, come back for doing it with MMA if you have issues or help $\endgroup$ – José Antonio Díaz Navas Dec 8 '17 at 18:21
