5.4

DLMF	Formula	Constraints	Maple	Mathematica	Symbolic Maple	Symbolic Mathematica	Numeric Maple	Numeric Mathematica
5.4#Ex1	${\displaystyle{\displaystyle\Gamma\left(1\right)=1}}$ \EulerGamma@{1} = 1		GAMMA(1) = 1	Gamma[1] == 1	Successful	Successful	-	Successful [Tested: 1]
5.4#Ex2	${\displaystyle{\displaystyle n!=\Gamma\left(n+1\right)}}$ n! = \EulerGamma@{n+1}	${\displaystyle{\displaystyle\Re(n+1)>0}}$	factorial(n) = GAMMA(n + 1)	(n)! == Gamma[n + 1]	Successful	Successful	-	Successful [Tested: 3]
5.4.E3	${\displaystyle{\displaystyle|\Gamma\left(iy\right)|=\left(\frac{\pi}{y\sinh\left(\pi y\right)}\right)^{1/2}}}$ |\EulerGamma@{iy}| = \left(\frac{\pi}{y\sinh@{\pi y}}\right)^{1/2}	${\displaystyle{\displaystyle\Re(\mathrm{i}y)>0}}$	abs(GAMMA(I*y)) = ((Pi)/(y*sinh(Pi*y)))^(1/2)	Abs[Gamma[I*y]] == (Divide[Pi,y*Sinh[Pi*y]])^(1/2)	Failure	Failure	Error	Skip - No test values generated
5.4.E4	${\displaystyle{\displaystyle\Gamma\left(\tfrac{1}{2}+\mathrm{i}y\right)\Gamma\left(\tfrac{1}{2}-\mathrm{i}y\right)=\left|\Gamma\left(\tfrac{1}{2}+\mathrm{i}y\right)\right|^{2}}}$ \EulerGamma@{\tfrac{1}{2}+\iunit y}\EulerGamma@{\tfrac{1}{2}-\iunit y} = \left|\EulerGamma@{\tfrac{1}{2}+\iunit y}\right|^{2}	${\displaystyle{\displaystyle\Re(\tfrac{1}{2}+\mathrm{i}y)>0,\Re(\tfrac{1}{2}-\mathrm{i}y)>0}}$	GAMMA((1)/(2)+ I*y)*GAMMA((1)/(2)- I*y) = (abs(GAMMA((1)/(2)+ I*y)))^(2)	Gamma[Divide[1,2]+ I*y]*Gamma[Divide[1,2]- I*y] == (Abs[Gamma[Divide[1,2]+ I*y]])^(2)	Failure	Failure	Successful [Tested: 6]	Successful [Tested: 6]
5.4.E4	${\displaystyle{\displaystyle\left|\Gamma\left(\tfrac{1}{2}+\mathrm{i}y\right)\right|^{2}=\frac{\pi}{\cosh\left(\pi y\right)}}}$ \left|\EulerGamma@{\tfrac{1}{2}+\iunit y}\right|^{2} = \frac{\pi}{\cosh@{\pi y}}	${\displaystyle{\displaystyle\Re(\tfrac{1}{2}+\mathrm{i}y)>0,\Re(\tfrac{1}{2}-\mathrm{i}y)>0}}$	(abs(GAMMA((1)/(2)+ I*y)))^(2) = (Pi)/(cosh(Pi*y))	(Abs[Gamma[Divide[1,2]+ I*y]])^(2) == Divide[Pi,Cosh[Pi*y]]	Failure	Failure	Successful [Tested: 6]	Successful [Tested: 6]
5.4.E5	${\displaystyle{\displaystyle\Gamma\left(\tfrac{1}{4}+\mathrm{i}y\right)\Gamma\left(\tfrac{3}{4}-\mathrm{i}y\right)=\frac{\pi\sqrt{2}}{\cosh\left(\pi y\right)+\mathrm{i}\sinh\left(\pi y\right)}}}$ \EulerGamma@{\tfrac{1}{4}+\iunit y}\EulerGamma@{\tfrac{3}{4}-\iunit y} = \frac{\pi\sqrt{2}}{\cosh@{\pi y}+\iunit\sinh@{\pi y}}	${\displaystyle{\displaystyle\Re(\tfrac{1}{4}+\mathrm{i}y)>0,\Re(\tfrac{3}{4}-\mathrm{i}y)>0}}$	GAMMA((1)/(4)+ I*y)*GAMMA((3)/(4)- I*y) = (Pi*sqrt(2))/(cosh(Pi*y)+ I*sinh(Pi*y))	Gamma[Divide[1,4]+ I*y]*Gamma[Divide[3,4]- I*y] == Divide[Pi*Sqrt[2],Cosh[Pi*y]+ I*Sinh[Pi*y]]	Successful	Successful	-	Successful [Tested: 6]
5.4.E6	${\displaystyle{\displaystyle\Gamma\left(\tfrac{1}{2}\right)=\pi^{1/2}\\ }}$ \EulerGamma@{\tfrac{1}{2}} = \pi^{1/2}\\		GAMMA((1)/(2)) = (Pi)^(1/2)	Gamma[Divide[1,2]] == (Pi)^(1/2)	Successful	Successful	-	Successful [Tested: 1]
5.4.E6	${\displaystyle{\displaystyle\pi^{1/2}\\ =1.77245\;38509\;05516\;02729\;\dots}}$ \pi^{1/2}\\ = 1.77245\;38509\;05516\;02729\;\dots		(Pi)^(1/2) = 1.77245385090551602729	(Pi)^(1/2) == 1.77245385090551602729	Successful	Successful	-	Successful [Tested: 1]
5.4.E7	${\displaystyle{\displaystyle\Gamma\left(\tfrac{1}{3}\right)=2.67893\;85347\;07747\;63365\;\dots}}$ \EulerGamma@{\tfrac{1}{3}} = 2.67893\;85347\;07747\;63365\;\dots		GAMMA((1)/(3)) = 2.67893853470774763365	Gamma[Divide[1,3]] == 2.67893853470774763365	Failure	Successful	Successful [Tested: 0]	Successful [Tested: 1]
5.4.E8	${\displaystyle{\displaystyle\Gamma\left(\tfrac{2}{3}\right)=1.35411\;79394\;26400\;41694\;\dots}}$ \EulerGamma@{\tfrac{2}{3}} = 1.35411\;79394\;26400\;41694\;\dots		GAMMA((2)/(3)) = 1.35411793942640041694	Gamma[Divide[2,3]] == 1.35411793942640041694	Successful	Successful	-	Successful [Tested: 1]
5.4.E9	${\displaystyle{\displaystyle\Gamma\left(\tfrac{1}{4}\right)=3.62560\;99082\;21908\;31193\;\dots}}$ \EulerGamma@{\tfrac{1}{4}} = 3.62560\;99082\;21908\;31193\;\dots		GAMMA((1)/(4)) = 3.62560990822190831193	Gamma[Divide[1,4]] == 3.62560990822190831193	Failure	Successful	Successful [Tested: 0]	Successful [Tested: 1]
5.4.E10	${\displaystyle{\displaystyle\Gamma\left(\tfrac{3}{4}\right)=1.22541\;67024\;65177\;64512\;\dots}}$ \EulerGamma@{\tfrac{3}{4}} = 1.22541\;67024\;65177\;64512\;\dots		GAMMA((3)/(4)) = 1.22541670246517764512	Gamma[Divide[3,4]] == 1.22541670246517764512	Successful	Successful	-	Successful [Tested: 1]
5.4.E11	${\displaystyle{\displaystyle\Gamma'\left(1\right)=-\gamma}}$ \EulerGamma'@{1} = -\EulerConstant		subs( temp=1, diff( GAMMA(temp), temp$(1) ) ) = - gamma	(D[Gamma[temp], {temp, 1}]/.temp-> 1) == - EulerGamma	Successful	Successful	-	Successful [Tested: 1]
5.4#Ex3	${\displaystyle{\displaystyle\psi\left(1\right)=-\gamma}}$ \digamma@{1} = -\EulerConstant		Psi(1) = - gamma	PolyGamma[1] == - EulerGamma	Successful	Successful	-	Successful [Tested: 1]
5.4#Ex4	${\displaystyle{\displaystyle\psi'\left(1\right)=\tfrac{1}{6}\pi^{2}}}$ \digamma'@{1} = \tfrac{1}{6}\pi^{2}		subs( temp=1, diff( Psi(temp), temp$(1) ) ) = (1)/(6)*(Pi)^(2)	(D[PolyGamma[temp], {temp, 1}]/.temp-> 1) == Divide[1,6]*(Pi)^(2)	Successful	Successful	-	Successful [Tested: 1]
5.4#Ex5	${\displaystyle{\displaystyle\psi\left(\tfrac{1}{2}\right)=-\gamma-2\ln 2}}$ \digamma@{\tfrac{1}{2}} = -\EulerConstant-2\ln@@{2}		Psi((1)/(2)) = - gamma - 2*ln(2)	PolyGamma[Divide[1,2]] == - EulerGamma - 2*Log[2]	Successful	Successful	-	Successful [Tested: 1]
5.4#Ex6	${\displaystyle{\displaystyle\psi'\left(\tfrac{1}{2}\right)=\tfrac{1}{2}\pi^{2}}}$ \digamma'@{\tfrac{1}{2}} = \tfrac{1}{2}\pi^{2}		subs( temp=(1)/(2), diff( Psi(temp), temp$(1) ) ) = (1)/(2)*(Pi)^(2)	(D[PolyGamma[temp], {temp, 1}]/.temp-> Divide[1,2]) == Divide[1,2]*(Pi)^(2)	Successful	Successful	-	Successful [Tested: 1]
5.4.E14	${\displaystyle{\displaystyle\psi\left(n+1\right)=\sum_{k=1}^{n}\frac{1}{k}-\gamma}}$ \digamma@{n+1} = \sum_{k=1}^{n}\frac{1}{k}-\EulerConstant		Psi(n + 1) = sum((1)/(k), k = 1..n)- gamma	PolyGamma[n + 1] == Sum[Divide[1,k], {k, 1, n}, GenerateConditions->None]- EulerGamma	Successful	Successful	-	Successful [Tested: 3]
5.4.E16	${\displaystyle{\displaystyle\Im\psi\left(iy\right)=\frac{1}{2y}+\frac{\pi}{2}\coth\left(\pi y\right)}}$ \imagpart@@{\digamma@{iy}} = \frac{1}{2y}+\frac{\pi}{2}\coth@{\pi y}		Im(Psi(I*y)) = (1)/(2*y)+(Pi)/(2)*coth(Pi*y)	Im[PolyGamma[I*y]] == Divide[1,2*y]+Divide[Pi,2]*Coth[Pi*y]	Failure	Failure	Successful [Tested: 6]	Successful [Tested: 6]
5.4.E17	${\displaystyle{\displaystyle\Im\psi\left(\tfrac{1}{2}+iy\right)=\frac{\pi}{2}\tanh\left(\pi y\right)}}$ \imagpart@@{\digamma@{\tfrac{1}{2}+iy}} = \frac{\pi}{2}\tanh@{\pi y}		Im(Psi((1)/(2)+ I*y)) = (Pi)/(2)*tanh(Pi*y)	Im[PolyGamma[Divide[1,2]+ I*y]] == Divide[Pi,2]*Tanh[Pi*y]	Failure	Failure	Successful [Tested: 6]	Successful [Tested: 6]
5.4.E18	${\displaystyle{\displaystyle\Im\psi\left(1+iy\right)=-\frac{1}{2y}+\frac{\pi}{2}\coth\left(\pi y\right)}}$ \imagpart@@{\digamma@{1+iy}} = -\frac{1}{2y}+\frac{\pi}{2}\coth@{\pi y}		Im(Psi(1 + I*y)) = -(1)/(2*y)+(Pi)/(2)*coth(Pi*y)	Im[PolyGamma[1 + I*y]] == -Divide[1,2*y]+Divide[Pi,2]*Coth[Pi*y]	Failure	Failure	Successful [Tested: 6]	Successful [Tested: 6]
5.4.E19	${\displaystyle{\displaystyle\psi\left(\frac{p}{q}\right)=-\gamma-\ln q-\frac{\pi}{2}\cot\left(\frac{\pi p}{q}\right)+\frac{1}{2}\sum_{k=1}^{q-1}\cos\left(\frac{2\pi kp}{q}\right)\ln\left(2-2\cos\left(\frac{2\pi k}{q}\right)\right)}}$ \digamma@{\frac{p}{q}} = -\EulerConstant-\ln@@{q}-\frac{\pi}{2}\cot@{\frac{\pi p}{q}}+\frac{1}{2}\sum_{k=1}^{q-1}\cos@{\frac{2\pi kp}{q}}\ln@{2-2\cos@{\frac{2\pi k}{q}}}		Psi((p)/(q)) = - gamma - ln(q)-(Pi)/(2)*cot((Pi*p)/(q))+(1)/(2)*sum(cos((2*Pi*k*p)/(q))*ln(2 - 2*cos((2*Pi*k)/(q))), k = 1..q - 1)	PolyGamma[Divide[p,q]] == - EulerGamma - Log[q]-Divide[Pi,2]*Cot[Divide[Pi*p,q]]+Divide[1,2]*Sum[Cos[Divide[2*Pi*k*p,q]]*Log[2 - 2*Cos[Divide[2*Pi*k,q]]], {k, 1, q - 1}, GenerateConditions->None]	Failure	Failure	Skipped - Because timed out	Skipped - Because timed out