5.6

DLMF	Formula	Constraints	Maple	Mathematica	Symbolic Maple	Symbolic Mathematica	Numeric Maple	Numeric Mathematica
5.6.E1	${\displaystyle{\displaystyle 1<(2\pi)^{-1/2}x^{(1/2)-x}e^{x}\Gamma\left(x\right)}}$ 1 < (2\pi)^{-1/2}x^{(1/2)-x}e^{x}\EulerGamma@{x}	${\displaystyle{\displaystyle\Re x>0}}$	1 < (2*Pi)^(- 1/2)* (x)^((1/2)- x)* exp(x)*GAMMA(x)	1 < (2*Pi)^(- 1/2)* (x)^((1/2)- x)* Exp[x]*Gamma[x]	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
5.6.E1	${\displaystyle{\displaystyle(2\pi)^{-1/2}x^{(1/2)-x}e^{x}\Gamma\left(x\right)<e^{1/(12x)}}}$ (2\pi)^{-1/2}x^{(1/2)-x}e^{x}\EulerGamma@{x} < e^{1/(12x)}	${\displaystyle{\displaystyle\Re x>0}}$	(2*Pi)^(- 1/2)* (x)^((1/2)- x)* exp(x)*GAMMA(x) < exp(1/(12*x))	(2*Pi)^(- 1/2)* (x)^((1/2)- x)* Exp[x]*Gamma[x] < Exp[1/(12*x)]	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
5.6.E2	${\displaystyle{\displaystyle\frac{1}{\Gamma\left(x\right)}+\frac{1}{\Gamma\left(1/x\right)}\leq 2}}$ \frac{1}{\EulerGamma@{x}}+\frac{1}{\EulerGamma@{1/x}} \leq 2	${\displaystyle{\displaystyle\Re x>0,\Re(1/x)>0}}$	(1)/(GAMMA(x))+(1)/(GAMMA(1/x)) <= 2	Divide[1,Gamma[x]]+Divide[1,Gamma[1/x]] <= 2	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
5.6.E3	${\displaystyle{\displaystyle\frac{1}{(\Gamma\left(x\right))^{2}}+\frac{1}{(\Gamma\left(1/x\right))^{2}}\leq 2}}$ \frac{1}{(\EulerGamma@{x})^{2}}+\frac{1}{(\EulerGamma@{1/x})^{2}} \leq 2	${\displaystyle{\displaystyle\Re x>0,\Re(1/x)>0}}$	(1)/((GAMMA(x))^(2))+(1)/((GAMMA(1/x))^(2)) <= 2	Divide[1,(Gamma[x])^(2)]+Divide[1,(Gamma[1/x])^(2)] <= 2	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
5.6.E4	${\displaystyle{\displaystyle\frac{\Gamma\left(x+1\right)}{\Gamma\left(x+s\right)}<(x+1)^{1-s}}}$ \frac{\EulerGamma@{x+1}}{\EulerGamma@{x+s}} < (x+1)^{1-s}	${\displaystyle{\displaystyle 0<s,s<1,\Re(x+1)>0,\Re(x+s)>0}}$	(GAMMA(x + 1))/(GAMMA(x + s)) < (x + 1)^(1 - s)	Divide[Gamma[x + 1],Gamma[x + s]] < (x + 1)^(1 - s)	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
5.6.E5	${\displaystyle{\displaystyle\exp\left((1-s)\psi\left(x+s^{1/2}\right)\right)\leq\frac{\Gamma\left(x+1\right)}{\Gamma\left(x+s\right)}}}$ \exp@{(1-s)\digamma@{x+s^{1/2}}} \leq \frac{\EulerGamma@{x+1}}{\EulerGamma@{x+s}}	${\displaystyle{\displaystyle 0<s,s<1,\Re(x+1)>0,\Re(x+s)>0}}$	exp((1 - s)*Psi(x + (s)^(1/2))) <= (GAMMA(x + 1))/(GAMMA(x + s))	Exp[(1 - s)*PolyGamma[x + (s)^(1/2)]] <= Divide[Gamma[x + 1],Gamma[x + s]]	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
5.6.E5	${\displaystyle{\displaystyle\frac{\Gamma\left(x+1\right)}{\Gamma\left(x+s\right)}\leq\exp\left((1-s)\psi\left(x+\tfrac{1}{2}(s+1)\right)\right)}}$ \frac{\EulerGamma@{x+1}}{\EulerGamma@{x+s}} \leq \exp@{(1-s)\digamma@{x+\tfrac{1}{2}(s+1)}}	${\displaystyle{\displaystyle 0<s,s<1,\Re(x+1)>0,\Re(x+s)>0}}$	(GAMMA(x + 1))/(GAMMA(x + s)) <= exp((1 - s)*Psi(x +(1)/(2)*(s + 1)))	Divide[Gamma[x + 1],Gamma[x + s]] <= Exp[(1 - s)*PolyGamma[x +Divide[1,2]*(s + 1)]]	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
5.6.E6	${\displaystyle{\displaystyle|\Gamma\left(x+\mathrm{i}y\right)|\leq|\Gamma\left(x\right)|}}$ |\EulerGamma@{x+\iunit y}| \leq |\EulerGamma@{x}|	${\displaystyle{\displaystyle\Re(x+\mathrm{i}y)>0,\Re x>0}}$	abs(GAMMA(x + I*y)) <= abs(GAMMA(x))	Abs[Gamma[x + I*y]] <= Abs[Gamma[x]]	Failure	Failure	Successful [Tested: 18]	Successful [Tested: 18]
5.6.E7	${\displaystyle{\displaystyle|\Gamma\left(x+\mathrm{i}y\right)|\geq(\operatorname{sech}\left(\pi y\right))^{1/2}\Gamma\left(x\right)}}$ |\EulerGamma@{x+\iunit y}| \geq (\sech@{\pi y})^{1/2}\EulerGamma@{x}	${\displaystyle{\displaystyle x\geq\tfrac{1}{2},\Re(x+\mathrm{i}y)>0,\Re x>0}}$	abs(GAMMA(x + I*y)) >= (sech(Pi*y))^(1/2)* GAMMA(x)	Abs[Gamma[x + I*y]] >= (Sech[Pi*y])^(1/2)* Gamma[x]	Failure	Failure	Successful [Tested: 18]	Successful [Tested: 18]
5.6.E8	${\displaystyle{\displaystyle\left|\frac{\Gamma\left(z+a\right)}{\Gamma\left(z+b\right)}\right|\leq\frac{1}{|z|^{b-a}}}}$ \left|\frac{\EulerGamma@{z+a}}{\EulerGamma@{z+b}}\right| \leq \frac{1}{|z|^{b-a}}	${\displaystyle{\displaystyle\Re(z+a)>0,\Re(z+b)>0}}$	abs((GAMMA(z + a))/(GAMMA(z + b))) <= (1)/((abs(z))^(b - a))	Abs[Divide[Gamma[z + a],Gamma[z + b]]] <= Divide[1,(Abs[z])^(b - a)]	Failure	Failure	Failed [30 / 83] Result: .5333333334 <= .1250000000 Test Values: {a = -1.5, b = 1.5, z = 2} Result: 2.000000000 <= .5000000000 Test Values: {a = -1.5, b = -.5, z = 2} Result: 1.333333334 <= .2500000000 Test Values: {a = -1.5, b = .5, z = 2} Result: .2954089752 <= .8838834764e-1 Test Values: {a = -1.5, b = 2, z = 2} ... skip entries to safe data	Failed [35 / 95] Result: False Test Values: {Rule[a, -1.5], Rule[b, 1.5], Rule[z, 2]} Result: False Test Values: {Rule[a, -1.5], Rule[b, -0.5], Rule[z, 2]} ... skip entries to safe data
5.6.E9	${\displaystyle{\displaystyle|\Gamma\left(z\right)|\leq(2\pi)^{1/2}|z|^{x-(1/2)}e^{-\pi|y|/2}\exp\left(\tfrac{1}{6}|z|^{-1}\right)}}$ |\EulerGamma@{z}| \leq (2\pi)^{1/2}|z|^{x-(1/2)}e^{-\pi|y|/2}\exp@{\tfrac{1}{6}|z|^{-1}}	${\displaystyle{\displaystyle\Re z>0}}$	abs(GAMMA(x + y*I)) <= (2*Pi)^(1/2)*(abs(x + y*I))^(x -(1/2))* exp(- Pi*abs(y)/2)*exp((1)/(6)*(abs(x + y*I))^(- 1))	Abs[Gamma[x + y*I]] <= (2*Pi)^(1/2)*(Abs[x + y*I])^(x -(1/2))* Exp[- Pi*Abs[y]/2]*Exp[Divide[1,6]*(Abs[x + y*I])^(- 1)]	Failure	Failure	Successful [Tested: 18]	Successful [Tested: 18]