5.12

DLMF	Formula	Constraints	Maple	Mathematica	Symbolic Maple	Symbolic Mathematica	Numeric Maple	Numeric Mathematica
5.12.E1	${\displaystyle{\displaystyle\mathrm{B}\left(a,b\right)=\int_{0}^{1}t^{a-1}(1-t)^{b-1}\mathrm{d}t}}$ \EulerBeta@{a}{b} = \int_{0}^{1}t^{a-1}(1-t)^{b-1}\diff{t}		Beta(a, b) = int((t)^(a - 1)*(1 - t)^(b - 1), t = 0..1)	Beta[a, b] == Integrate[(t)^(a - 1)*(1 - t)^(b - 1), {t, 0, 1}, GenerateConditions->None]	Failure	Successful	Error	Failed [11 / 36] Result: Indeterminate Test Values: {Rule[a, -1.5], Rule[b, -2]} Result: Indeterminate Test Values: {Rule[a, 1.5], Rule[b, -2]} ... skip entries to safe data
5.12.E1	${\displaystyle{\displaystyle\int_{0}^{1}t^{a-1}(1-t)^{b-1}\mathrm{d}t=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}}}$ \int_{0}^{1}t^{a-1}(1-t)^{b-1}\diff{t} = \frac{\EulerGamma@{a}\EulerGamma@{b}}{\EulerGamma@{a+b}}	${\displaystyle{\displaystyle\Re a>0,\Re b>0,\Re(a+b)>0}}$	int((t)^(a - 1)*(1 - t)^(b - 1), t = 0..1) = (GAMMA(a)*GAMMA(b))/(GAMMA(a + b))	Integrate[(t)^(a - 1)*(1 - t)^(b - 1), {t, 0, 1}, GenerateConditions->None] == Divide[Gamma[a]*Gamma[b],Gamma[a + b]]	Successful	Successful	Skip - symbolical successful subtest	Successful [Tested: 9]
5.12.E2	${\displaystyle{\displaystyle\int_{0}^{\pi/2}{\sin^{2a-1}}\theta{\cos^{2b-1}}\theta\mathrm{d}\theta=\tfrac{1}{2}\mathrm{B}\left(a,b\right)}}$ \int_{0}^{\pi/2}\sin^{2a-1}@@{\theta}\cos^{2b-1}@@{\theta}\diff{\theta} = \tfrac{1}{2}\EulerBeta@{a}{b}	${\displaystyle{\displaystyle\Re a>0,\Re b>0,\Re(a+b)>0}}$	int((sin(theta))^(2*a - 1)* (cos(theta))^(2*b - 1), theta = 0..Pi/2) = (1)/(2)*Beta(a, b)	Integrate[(Sin[\[Theta]])^(2*a - 1)* (Cos[\[Theta]])^(2*b - 1), {\[Theta], 0, Pi/2}, GenerateConditions->None] == Divide[1,2]*Beta[a, b]	Failure	Successful	Error	Successful [Tested: 9]
5.12.E3	${\displaystyle{\displaystyle\int_{0}^{\infty}\frac{t^{a-1}\mathrm{d}t}{(1+t)^{a+b}}=\mathrm{B}\left(a,b\right)}}$ \int_{0}^{\infty}\frac{t^{a-1}\diff{t}}{(1+t)^{a+b}} = \EulerBeta@{a}{b}	${\displaystyle{\displaystyle\Re a>0,\Re b>0,\Re(a+b)>0}}$	int(((t)^(a - 1))/((1 + t)^(a + b)), t = 0..infinity) = Beta(a, b)	Integrate[Divide[(t)^(a - 1),(1 + t)^(a + b)], {t, 0, Infinity}, GenerateConditions->None] == Beta[a, b]	Failure	Successful	Error	Successful [Tested: 9]
5.12.E4	${\displaystyle{\displaystyle\int_{0}^{1}\frac{t^{a-1}(1-t)^{b-1}}{(t+z)^{a+b}}\mathrm{d}t=\mathrm{B}\left(a,b\right)(1+z)^{-a}z^{-b}}}$ \int_{0}^{1}\frac{t^{a-1}(1-t)^{b-1}}{(t+z)^{a+b}}\diff{t} = \EulerBeta@{a}{b}(1+z)^{-a}z^{-b}	${\displaystyle{\displaystyle|\operatorname{ph}z|<\pi,\Re a>0,\Re b>0,\Re(a+b)>0}}$	int(((t)^(a - 1)*(1 - t)^(b - 1))/((t + z)^(a + b)), t = 0..1) = Beta(a, b)*(1 + z)^(- a)* (z)^(- b)	Integrate[Divide[(t)^(a - 1)*(1 - t)^(b - 1),(t + z)^(a + b)], {t, 0, 1}, GenerateConditions->None] == Beta[a, b]*(1 + z)^(- a)* (z)^(- b)	Failure	Failure	Skipped - Because timed out	Failed [77 / 252] Result: Indeterminate Test Values: {Rule[a, -1.5], Rule[b, -2], Rule[z, Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]} Result: Indeterminate Test Values: {Rule[a, -1.5], Rule[b, -2], Rule[z, Power[E, Times[Complex[0, Rational[2, 3]], Pi]]]} ... skip entries to safe data
5.12.E5	${\displaystyle{\displaystyle\int_{0}^{\pi/2}(\cos t)^{a-1}\cos\left(bt\right)\mathrm{d}t=\frac{\pi}{2^{a}}\frac{1}{a\mathrm{B}\left(\frac{1}{2}(a+b+1),\frac{1}{2}(a-b+1)\right)}}}$ \int_{0}^{\pi/2}(\cos@@{t})^{a-1}\cos@{bt}\diff{t} = \frac{\pi}{2^{a}}\frac{1}{a\EulerBeta@{\frac{1}{2}(a+b+1)}{\frac{1}{2}(a-b+1)}}	${\displaystyle{\displaystyle\Re a>0,\Re(\frac{1}{2}(a+b+1))>0,\Re(\frac{1}{2}(a-b+1))>0,\Re((\frac{1}{2}(a+b+1))+b)>0,\Re(a+(\frac{1}{2}(a-b+1)))>0}}$	int((cos(t))^(a - 1)* cos(b*t), t = 0..Pi/2) = (Pi)/((2)^(a))*(1)/(a*Beta((1)/(2)*(a + b + 1), (1)/(2)*(a - b + 1)))	Integrate[(Cos[t])^(a - 1)* Cos[b*t], {t, 0, Pi/2}, GenerateConditions->None] == Divide[Pi,(2)^(a)]*Divide[1,a*Beta[Divide[1,2]*(a + b + 1), Divide[1,2]*(a - b + 1)]]	Failure	Aborted	Successful [Tested: 18]	Skipped - Because timed out
5.12.E6	${\displaystyle{\displaystyle\int_{0}^{\pi}(\sin t)^{a-1}e^{ibt}\mathrm{d}t=\frac{\pi}{2^{a-1}}\frac{e^{i\pi b/2}}{a\mathrm{B}\left(\frac{1}{2}(a+b+1),\frac{1}{2}(a-b+1)\right)}}}$ \int_{0}^{\pi}(\sin@@{t})^{a-1}e^{ibt}\diff{t} = \frac{\pi}{2^{a-1}}\frac{e^{i\pi b/2}}{a\EulerBeta@{\frac{1}{2}(a+b+1)}{\frac{1}{2}(a-b+1)}}	${\displaystyle{\displaystyle\Re a>0,\Re(\frac{1}{2}(a+b+1))>0,\Re(\frac{1}{2}(a-b+1))>0,\Re((\frac{1}{2}(a+b+1))+b)>0,\Re(a+(\frac{1}{2}(a-b+1)))>0}}$	int((sin(t))^(a - 1)* exp(I*b*t), t = 0..Pi) = (Pi)/((2)^(a - 1))*(exp(I*Pi*b/2))/(a*Beta((1)/(2)*(a + b + 1), (1)/(2)*(a - b + 1)))	Integrate[(Sin[t])^(a - 1)* Exp[I*b*t], {t, 0, Pi}, GenerateConditions->None] == Divide[Pi,(2)^(a - 1)]*Divide[Exp[I*Pi*b/2],a*Beta[Divide[1,2]*(a + b + 1), Divide[1,2]*(a - b + 1)]]	Failure	Aborted	Successful [Tested: 18]	Failed [9 / 18] Result: DirectedInfinity[] Test Values: {Rule[a, 1.5], Rule[b, -1.5]} Result: DirectedInfinity[] Test Values: {Rule[a, 1.5], Rule[b, 0.5]} ... skip entries to safe data
5.12.E7	${\displaystyle{\displaystyle\int_{0}^{\infty}\frac{\cosh\left(2bt\right)}{(\cosh t)^{2a}}\mathrm{d}t=4^{a-1}\mathrm{B}\left(a+b,a-b\right)}}$ \int_{0}^{\infty}\frac{\cosh@{2bt}}{(\cosh@@{t})^{2a}}\diff{t} = 4^{a-1}\EulerBeta@{a+b}{a-b}	${\displaystyle{\displaystyle\Re a>|\Re b|,\Re(a+b)>0,\Re(a-b)>0,\Re((a+b)+b)>0,\Re(a+(a-b))>0}}$	int((cosh(2*b*t))/((cosh(t))^(2*a)), t = 0..infinity) = (4)^(a - 1)* Beta(a + b, a - b)	Integrate[Divide[Cosh[2*b*t],(Cosh[t])^(2*a)], {t, 0, Infinity}, GenerateConditions->None] == (4)^(a - 1)* Beta[a + b, a - b]	Failure	Failure	Successful [Tested: 6]	Successful [Tested: 6]
5.12.E8	${\displaystyle{\displaystyle\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\mathrm{d}t}{(w+it)^{a}(z-it)^{b}}=\frac{(w+z)^{1-a-b}}{(a+b-1)\mathrm{B}\left(a,b\right)}}}$ \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\diff{t}}{(w+it)^{a}(z-it)^{b}} = \frac{(w+z)^{1-a-b}}{(a+b-1)\EulerBeta@{a}{b}}	${\displaystyle{\displaystyle\Re\left(a+b\right)>1,\Re w>0,\Re z>0,\Re a>0,\Re b>0,\Re(a+b)>0}}$	(1)/(2*Pi)*int((1)/((w + I*t)^(a)*(z - I*t)^(b)), t = - infinity..infinity) = ((w + z)^(1 - a - b))/((a + b - 1)*Beta(a, b))	Divide[1,2*Pi]*Integrate[Divide[1,(w + I*t)^(a)*(z - I*t)^(b)], {t, - Infinity, Infinity}, GenerateConditions->None] == Divide[(w + z)^(1 - a - b),(a + b - 1)*Beta[a, b]]	Skipped - Unable to analyze test case: Null	Failure	-	Successful [Tested: 250]
5.12.E9	${\displaystyle{\displaystyle\frac{1}{2\pi i}\int_{c-\infty i}^{c+\infty i}t^{-a}(1-t)^{-1-b}\mathrm{d}t=\frac{1}{b\mathrm{B}\left(a,b\right)}}}$ \frac{1}{2\pi i}\int_{c-\infty i}^{c+\infty i}t^{-a}(1-t)^{-1-b}\diff{t} = \frac{1}{b\EulerBeta@{a}{b}}	${\displaystyle{\displaystyle 0<c,c<1,\Re\left(a+b\right)>0,\Re a>0,\Re b>0,\Re(a+b)>0}}$	(1)/(2*Pi*I)*int((t)^(- a)*(1 - t)^(- 1 - b), t = c - infinity*I..c + infinity*I) = (1)/(b*Beta(a, b))	Divide[1,2*Pi*I]*Integrate[(t)^(- a)*(1 - t)^(- 1 - b), {t, c - Infinity*I, c + Infinity*I}, GenerateConditions->None] == Divide[1,b*Beta[a, b]]	Skipped - Unable to analyze test case: Null	Aborted	-	Skipped - Because timed out
5.12.E10	${\displaystyle{\displaystyle\frac{1}{2\pi i}\int_{0}^{(1+)}t^{a-1}(t-1)^{b-1}\mathrm{d}t=\frac{\sin\left(\pi b\right)}{\pi}\mathrm{B}\left(a,b\right)}}$ \frac{1}{2\pi i}\int_{0}^{(1+)}t^{a-1}(t-1)^{b-1}\diff{t} = \frac{\sin@{\pi b}}{\pi}\EulerBeta@{a}{b}	${\displaystyle{\displaystyle\Re a>0,\Re b>0,\Re(a+b)>0}}$	(1)/(2*Pi*I)*int((t)^(a - 1)*(t - 1)^(b - 1), t = 0..(1 +)) = (sin(Pi*b))/(Pi)*Beta(a, b)	Divide[1,2*Pi*I]*Integrate[(t)^(a - 1)*(t - 1)^(b - 1), {t, 0, (1 +)}, GenerateConditions->None] == Divide[Sin[Pi*b],Pi]*Beta[a, b]	Skipped - Unable to analyze test case: Null	Failure	-	Error
5.12.E11	${\displaystyle{\displaystyle\frac{1}{e^{2\pi ia}-1}\int_{\infty}^{(0+)}t^{a-1}(1+t)^{-a-b}\mathrm{d}t=\mathrm{B}\left(a,b\right)}}$ \frac{1}{e^{2\pi ia}-1}\int_{\infty}^{(0+)}t^{a-1}(1+t)^{-a-b}\diff{t} = \EulerBeta@{a}{b}	${\displaystyle{\displaystyle\Re a>0,\Re b>0,\Re(a+b)>0}}$	(1)/(exp(2*Pi*I*a)- 1)*int((t)^(a - 1)*(1 + t)^(- a - b), t = infinity..(0 +)) = Beta(a, b)	Divide[1,Exp[2*Pi*I*a]- 1]*Integrate[(t)^(a - 1)*(1 + t)^(- a - b), {t, Infinity, (0 +)}, GenerateConditions->None] == Beta[a, b]	Error	Failure	-	Error
5.12.E12	${\displaystyle{\displaystyle\int_{P}^{(1+,0+,1-,0-)}t^{a-1}(1-t)^{b-1}\mathrm{d}t=-4e^{\pi i(a+b)}\sin\left(\pi a\right)\sin\left(\pi b\right)\mathrm{B}\left(a,b\right)}}$ \int_{P}^{(1+,0+,1-,0-)}t^{a-1}(1-t)^{b-1}\diff{t} = -4e^{\pi i(a+b)}\sin@{\pi a}\sin@{\pi b}\EulerBeta@{a}{b}	${\displaystyle{\displaystyle\Re a>0,\Re b>0,\Re(a+b)>0}}$	int((t)^(a - 1)*(1 - t)^(b - 1), t = P..(1 + , 0 + , 1 - , 0 -)) = - 4*exp(Pi*I*(a + b))*sin(Pi*a)*sin(Pi*b)*Beta(a, b)	Integrate[(t)^(a - 1)*(1 - t)^(b - 1), {t, P, (1 + , 0 + , 1 - , 0 -)}, GenerateConditions->None] == - 4*Exp[Pi*I*(a + b)]*Sin[Pi*a]*Sin[Pi*b]*Beta[a, b]	Error	Failure	-	Error