4.10

DLMF	Formula	Constraints	Maple	Mathematica	Symbolic Maple	Symbolic Mathematica	Numeric Maple	Numeric Mathematica
4.10.E1	${\displaystyle{\displaystyle\int\frac{\mathrm{d}z}{z}=\ln z}}$ \int\frac{\diff{z}}{z} = \ln@@{z}		int((1)/(z), z) = ln(z)	Integrate[Divide[1,z], z, GenerateConditions->None] == Log[z]	Successful	Successful	-	Successful [Tested: 7]
4.10.E2	${\displaystyle{\displaystyle\int\ln z\mathrm{d}z=z\ln z-z}}$ \int\ln@@{z}\diff{z} = z\ln@@{z}-z		int(ln(z), z) = z*ln(z)- z	Integrate[Log[z], z, GenerateConditions->None] == z*Log[z]- z	Successful	Successful	-	Successful [Tested: 7]
4.10.E3	${\displaystyle{\displaystyle\int z^{n}\ln z\mathrm{d}z=\frac{z^{n+1}}{n+1}\ln z-\frac{z^{n+1}}{(n+1)^{2}}}}$ \int z^{n}\ln@@{z}\diff{z} = \frac{z^{n+1}}{n+1}\ln@@{z}-\frac{z^{n+1}}{(n+1)^{2}}	${\displaystyle{\displaystyle n\neq-1}}$	int((z)^(n)* ln(z), z) = ((z)^(n + 1))/(n + 1)*ln(z)-((z)^(n + 1))/((n + 1)^(2))	Integrate[(z)^(n)* Log[z], z, GenerateConditions->None] == Divide[(z)^(n + 1),n + 1]*Log[z]-Divide[(z)^(n + 1),(n + 1)^(2)]	Successful	Successful	-	Successful [Tested: 21]
4.10.E4	${\displaystyle{\displaystyle\int\frac{\mathrm{d}z}{z\ln z}=\ln\left(\ln z\right)}}$ \int\frac{\diff{z}}{z\ln@@{z}} = \ln@{\ln@@{z}}		int((1)/(z*ln(z)), z) = ln(ln(z))	Integrate[Divide[1,z*Log[z]], z, GenerateConditions->None] == Log[Log[z]]	Successful	Successful	-	Successful [Tested: 7]
4.10.E5	${\displaystyle{\displaystyle\int_{0}^{1}\frac{\ln t}{1-t}\mathrm{d}t=-\frac{\pi^{2}}{6}}}$ \int_{0}^{1}\frac{\ln@@{t}}{1-t}\diff{t} = -\frac{\pi^{2}}{6}		int((ln(t))/(1 - t), t = 0..1) = -((Pi)^(2))/(6)	Integrate[Divide[Log[t],1 - t], {t, 0, 1}, GenerateConditions->None] == -Divide[(Pi)^(2),6]	Successful	Successful	-	Successful [Tested: 1]
4.10.E6	${\displaystyle{\displaystyle\int_{0}^{1}\frac{\ln t}{1+t}\mathrm{d}t=-\frac{\pi^{2}}{12}}}$ \int_{0}^{1}\frac{\ln@@{t}}{1+t}\diff{t} = -\frac{\pi^{2}}{12}		int((ln(t))/(1 + t), t = 0..1) = -((Pi)^(2))/(12)	Integrate[Divide[Log[t],1 + t], {t, 0, 1}, GenerateConditions->None] == -Divide[(Pi)^(2),12]	Successful	Successful	-	Successful [Tested: 1]
4.10.E8	${\displaystyle{\displaystyle\int e^{az}\mathrm{d}z=\frac{e^{az}}{a}}}$ \int e^{az}\diff{z} = \frac{e^{az}}{a}		int(exp(a*z), z) = (exp(a*z))/(a)	Integrate[Exp[a*z], z, GenerateConditions->None] == Divide[Exp[a*z],a]	Successful	Successful	-	Successful [Tested: 42]
4.10.E9	${\displaystyle{\displaystyle\int\frac{\mathrm{d}z}{e^{az}+b}=\frac{1}{ab}(az-\ln\left(e^{az}+b\right))}}$ \int\frac{\diff{z}}{e^{az}+b} = \frac{1}{ab}(az-\ln@{e^{az}+b})		int((1)/(exp(a*z)+ b), z) = (1)/(a*b)*(a*z - ln(exp(a*z)+ b))	Integrate[Divide[1,Exp[a*z]+ b], z, GenerateConditions->None] == Divide[1,a*b]*(a*z - Log[Exp[a*z]+ b])	Failure	Successful	Successful [Tested: 252]	Successful [Tested: 252]
4.10.E10	${\displaystyle{\displaystyle\int\frac{e^{az}-1}{e^{az}+1}\mathrm{d}z=\frac{2}{a}\ln\left(e^{az/2}+e^{-az/2}\right)}}$ \int\frac{e^{az}-1}{e^{az}+1}\diff{z} = \frac{2}{a}\ln@{e^{az/2}+e^{-az/2}}		int((exp(a*z)- 1)/(exp(a*z)+ 1), z) = (2)/(a)*ln(exp(a*z/2)+ exp(- a*z/2))	Integrate[Divide[Exp[a*z]- 1,Exp[a*z]+ 1], z, GenerateConditions->None] == Divide[2,a]*Log[Exp[a*z/2]+ Exp[- a*z/2]]	Failure	Failure	Successful [Tested: 42]	Successful [Tested: 42]
4.10.E11	${\displaystyle{\displaystyle\int_{-\infty}^{\infty}e^{-cx^{2}}\mathrm{d}x=\sqrt{\frac{\pi}{c}}}}$ \int_{-\infty}^{\infty}e^{-cx^{2}}\diff{x} = \sqrt{\frac{\pi}{c}}	${\displaystyle{\displaystyle\Re c>0}}$	int(exp(- c*(x)^(2)), x = - infinity..infinity) = sqrt((Pi)/(c))	Integrate[Exp[- c*(x)^(2)], {x, - Infinity, Infinity}, GenerateConditions->None] == Sqrt[Divide[Pi,c]]	Successful	Successful	-	Successful [Tested: 3]
4.10.E12	${\displaystyle{\displaystyle\int_{0}^{\ln 2}\frac{xe^{x}}{e^{x}-1}\mathrm{d}x=\frac{\pi^{2}}{12}}}$ \int_{0}^{\ln@@{2}}\frac{xe^{x}}{e^{x}-1}\diff{x} = \frac{\pi^{2}}{12}		int((x*exp(x))/(exp(x)- 1), x = 0..ln(2)) = ((Pi)^(2))/(12)	Integrate[Divide[x*Exp[x],Exp[x]- 1], {x, 0, Log[2]}, GenerateConditions->None] == Divide[(Pi)^(2),12]	Successful	Successful	-	Successful [Tested: 1]
4.10.E13	${\displaystyle{\displaystyle\int_{0}^{\infty}\frac{\mathrm{d}x}{e^{x}+1}=\ln 2}}$ \int_{0}^{\infty}\frac{\diff{x}}{e^{x}+1} = \ln@@{2}		int((1)/(exp(x)+ 1), x = 0..infinity) = ln(2)	Integrate[Divide[1,Exp[x]+ 1], {x, 0, Infinity}, GenerateConditions->None] == Log[2]	Successful	Successful	-	Successful [Tested: 1]