Gamma Function - 5.8 Infinite Products

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5.8.E2 1 Γ ( z ) = z e γ z k = 1 ( 1 + z k ) e - z / k 1 Euler-Gamma 𝑧 𝑧 superscript 𝑒 𝑧 superscript subscript product 𝑘 1 1 𝑧 𝑘 superscript 𝑒 𝑧 𝑘 {\displaystyle{\displaystyle\frac{1}{\Gamma\left(z\right)}=ze^{\gamma z}\prod_% {k=1}^{\infty}\left(1+\frac{z}{k}\right)e^{-z/k}}}
\frac{1}{\EulerGamma@{z}} = ze^{\EulerConstant z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)e^{-z/k}
z > 0 𝑧 0 {\displaystyle{\displaystyle\Re z>0}}
(1)/(GAMMA(z)) = z*exp(gamma*z)*product((1 +(z)/(k))*exp(- z/k), k = 1..infinity)
Divide[1,Gamma[z]] == z*Exp[EulerGamma*z]*Product[(1 +Divide[z,k])*Exp[- z/k], {k, 1, Infinity}, GenerateConditions->None]
Successful Successful - Successful [Tested: 5]
5.8.E3 | Γ ( x ) Γ ( x + i y ) | 2 = k = 0 ( 1 + y 2 ( x + k ) 2 ) superscript Euler-Gamma 𝑥 Euler-Gamma 𝑥 imaginary-unit 𝑦 2 superscript subscript product 𝑘 0 1 superscript 𝑦 2 superscript 𝑥 𝑘 2 {\displaystyle{\displaystyle\left|\frac{\Gamma\left(x\right)}{\Gamma\left(x+% \mathrm{i}y\right)}\right|^{2}=\prod_{k=0}^{\infty}\left(1+\frac{y^{2}}{(x+k)^% {2}}\right)}}
\left|\frac{\EulerGamma@{x}}{\EulerGamma@{x+\iunit y}}\right|^{2} = \prod_{k=0}^{\infty}\left(1+\frac{y^{2}}{(x+k)^{2}}\right)
x > 0 , ( x + i y ) > 0 formulae-sequence 𝑥 0 𝑥 imaginary-unit 𝑦 0 {\displaystyle{\displaystyle\Re x>0,\Re(x+\mathrm{i}y)>0}}
(abs((GAMMA(x))/(GAMMA(x + I*y))))^(2) = product(1 +((y)^(2))/((x + k)^(2)), k = 0..infinity)
(Abs[Divide[Gamma[x],Gamma[x + I*y]]])^(2) == Product[1 +Divide[(y)^(2),(x + k)^(2)], {k, 0, Infinity}, GenerateConditions->None]
Failure Failure
Failed [18 / 18]
Result: -6.243891895+0.*I
Test Values: {x = 1.5, y = -1.5, x = 1}

Result: -6.243891895+0.*I
Test Values: {x = 1.5, y = 1.5, x = 1}

Result: -.210463144+0.*I
Test Values: {x = 1.5, y = -.5, x = 1}

Result: -.210463144+0.*I
Test Values: {x = 1.5, y = .5, x = 1}

... skip entries to safe data
Successful [Tested: 18]