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e-z-1-f-1-a-b-z-b-b-a-G-1-2-1-1-z-0-1-b-a-b-1-




Question Number 88417 by M±th+et£s last updated on 10/Apr/20
e^(−z)  _1 f_1 (a;b;z)=((Γ(b))/(Γ(b−a))) G_(1,2) ^(1,1) (z∣_(0,1−b) ^(a−b+1) )
$${e}^{−{z}} \:_{\mathrm{1}} {f}_{\mathrm{1}} \left({a};{b};{z}\right)=\frac{\Gamma\left({b}\right)}{\Gamma\left({b}−{a}\right)}\:{G}_{\mathrm{1},\mathrm{2}} ^{\mathrm{1},\mathrm{1}} \left({z}\mid_{\mathrm{0},\mathrm{1}−{b}} ^{{a}−{b}+\mathrm{1}} \right) \\ $$
Commented by M±th+et£s last updated on 10/Apr/20
G is meijer function  f is hypergeometric function
$${G}\:{is}\:{meijer}\:{function} \\ $$$${f}\:{is}\:{hypergeometric}\:{function} \\ $$

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