Integrate modified in version 6?
- To: mathgroup at smc.vnet.net
- Subject: [mg77938] Integrate modified in version 6?
- From: dimitris <dimmechan at yahoo.com>
- Date: Wed, 20 Jun 2007 05:30:14 -0400 (EDT)
I don't have 6 to be more rigorous but based on some integrals post in another forum by Vladimir Bondarenko I am quite sure that something has change in the integration algorithm for definite integrals... Say for example the integral Integrate[z ArcSin[z]/(1+z)^2, {z, 0, 1}]. In Mathematica 6 we have Integrate[z ArcSin[z]/(1+z)^2, {z, 0, 1}] -Infinity which is far away from truth. The good old (?) Mathematica 5.2 returns In[3]:= Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}] {N[%], NIntegrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]} Out[3]= -1 - 2*Catalan + Pi*(1/4 + Log[2]) Out[4]= {0.13105306534661265, 0.1310530653479215} Let's add an rule for Limit in 5.2 Unprotect[Limit]; Limit[x___] := Null /; Print[InputForm[limit[x]]]; Let's get the integral In[4]:= Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}] >From In[4]:= limit[z^2, z -> 0, Direction -> -1, Assumptions -> True] >From In[4]:= limit[(1/4 - (-1 + z)^2/16)*(Pi/2 + I*Sqrt[2]*Sqrt[-1 + z] - ((I/6)*(-1 + z)^(3/2))/Sqrt[2] + (((3*I)/80)*(-1 + z)^(5/2))/ Sqrt[2]), z -> 1, Direction -> 1, Assumptions -> True] >From In[4]:= limit[(1/4 - (-1 + z)^2/16)*(Pi/2 + I*Sqrt[2]*Sqrt[-1 + z] - ((I/6)*(-1 + z)^(3/2))/Sqrt[2] + (((3*I)/80)*(-1 + z)^(5/2))/ Sqrt[2]), z -> 1, Direction -> 1, Assumptions -> True] >From In[4]:= limit[z^2, z -> 0, Direction -> -1, Assumptions -> True] >From In[4]:= limit[((-I)*(1 + z)*ArcSin[z]^2 + ArcSin[z]*(2 + I*Pi*(1 + z) + 4*( 1 + z)*Log[1 - I*E^( I*ArcSin[z])]) + 2*(Sqrt[1 - z^2] + Pi*( 1 + z)*Log[1 - I*E^( I*ArcSin[z])] + 2*Pi*(1 + z)*Log[(1 + E^(I*ArcSin[z]))/ E^(I* ArcSin[z])] - 2*Pi*Log[Cos[ArcSin[z]/2]] - 2* Pi*z*Log[Cos[ArcSin[z]/ 2]] - Pi*Log[Sin[(Pi + 2*ArcSin[z])/4]] - Pi*z*Log[ Sin[(Pi + 2*ArcSin[z])/4]]) - (4* I)*(1 + z)*PolyLog[2, I*E^(I*ArcSin[z])])/(2*(1 + z)), z -> 1, Direction -> 1, Assumptions -> True] >From In[4]:= limit[((-I)*(1 + z)*ArcSin[z]^2 + ArcSin[z]*(2 + I*Pi*(1 + z) + 4*( 1 + z)*Log[1 - I*E^( I*ArcSin[z])]) + 2*(Sqrt[1 - z^2] + Pi*( 1 + z)*Log[1 - I*E^( I*ArcSin[z])] + 2*Pi*(1 + z)*Log[(1 + E^(I*ArcSin[z]))/ E^(I* ArcSin[z])] - 2*Pi*Log[Cos[ArcSin[z]/2]] - 2* Pi*z*Log[Cos[ArcSin[z]/ 2]] - Pi*Log[Sin[(Pi + 2*ArcSin[z])/4]] - Pi*z*Log[ Sin[(Pi + 2*ArcSin[z])/4]]) - (4* I)*(1 + z)*PolyLog[2, I*E^(I*ArcSin[z])])/(2*(1 + z)), z -> 0, Direction -> -1, Assumptions -> True] It can be seen that the integral is evaluated by application of the NL formula. Let's do the same in version 6: Here is the output as Vladimir sent me (sortening a little!) limit[z^2, z -> 0, Direction -> -1, Assumptions -> True] limit[Pi/8 - ((I/2)*Sqrt[-1 + z])/Sqrt[2] + ((I/24)*(-1 + \ z)^(3/2))/Sqrt[2] - (Pi*(-1 + z)^2)/32, z -> 1, Direction -> 1, \ Assumptions -> True] limit[Pi/8 - ((I/2)*Sqrt[-1 + z])/Sqrt[2] + ((I/24)*(-1 + \ z)^(3/2))/Sqrt[2] - (Pi*(-1 + z)^2)/32, z -> 1, Direction -> 1, \ Assumptions -> True] limit[z^2, z -> 0, Direction -> -1, Assumptions -> True] limit[z*((Pi*Integrate`ImproperDump`MeijerGfunction[{}, {1}, {0}, {}, \ z^2])/2 - (Sqrt[Pi]*Integrate`ImproperDump`MeijerGfunction[{}, {1, \ 1}, {0, 1/2}, {}, \ z^2])/2)*Integrate`ImproperDump`MeijerGfunction[{-1}, {}, {0}, {}, \ z], z -> Infinity, Assumptions -> True] limit[2/(z^2*(Pi*Integrate`ImproperDump`MeijerGfunction[{}, {1}, {0}, \ {}, z^2] - Sqrt[Pi]*Integrate`ImproperDump`MeijerGfunction[{}, {1, \ 1}, {0, 1/2}, {}, z^2])*Integrate`ImproperDump`MeijerGfunction[{-1}, \ {}, {0}, {}, z]), z -> Infinity, Assumptions -> True] limit[System`MeijerGDump`zz$2982, System`MeijerGDump`zz$2982 -> 1] limit[(-3*Pi*System`MeijerGDump`zz$2982)/2, \ System`MeijerGDump`zz$2982 -> 1] limit[(Pi*(-Sqrt[System`MeijerGDump`zz$2982] - \ ArcTan[Sqrt[System`MeijerGDump`zz$2982]] + \ System`MeijerGDump`zz$2982*ArcTan[Sqrt[System`MeijerGDump`zz $2982]]))/ \ (-1 + System`MeijerGDump`zz$2982) - \ (3*Pi*System`MeijerGDump`zz$2982*((2*System`MeijerGDump`zz$2982*((1 - \ System`MeijerGDump`zz$2982)^(-1) - (-System`MeijerGDump`zz$2982 - \ Log[1 - System`MeijerGDump`zz$2982])/System`MeijerGDump`zz$2982^2))/3 \ - Log[1 - System`MeijerGDump`zz$2982]/System`MeijerGDump`zz$2982))/2 \ + (Pi*(-Log[System`MeijerGDump`zz$2982] + PolyGamma[0, 1/2] - \ PolyGamma[0, 3/2]))/2, System`MeijerGDump`zz$2982 -> 1, Direction -> \ 1] limit[(2*K$3121*(1 + K$3121)*System`MeijerGDump`zz$3095^(1/2 + \ K$3121)*Gamma[1/2 + K$3121])/((1 + 2*K$3121)*Gamma[1 + K$3121]), \ K$3121 -> Infinity, Assumptions -> True] limit[System`MeijerGDump`zz$3095, K$3121 -> Infinity, Analytic -> \ True, Assumptions -> K$3121 > 1073741824] limit[K$3121^(-1), K$3121 -> 0, Assumptions -> K$3121^(-1) > \ 1073741824] limit[K$3121^(-1), K$3121 -> 0, Assumptions -> K$3121^(-1) > \ 1073741824] limit[(2 + K$3121)/(2*K$3121), K$3121 -> 0, Assumptions -> \ K$3121^(-1) > 1073741824] limit[(2 + K$3121)/(2*K$3121), K$3121 -> 0, Assumptions -> \ K$3121^(-1) > 1073741824] limit[((2 + K$3121)*(1 + \ 2*K$3121)^2*System`MeijerGDump`zz$3095)/(2*(1 + K$3121)^2*(3 + \ 2*K$3121)), K$3121 -> Infinity, Assumptions -> True] limit[System`MeijerGDump`zz$3095^(1/2 + K$3806)/Sqrt[K$3806], K$3806 - > \ Infinity, Assumptions -> True] -Infinity Bang! >From this output I am quite sure that no NL thoerem to the indefinite integral takes place but rather straightly convolution (I write it correct now; Cheers David Cantrell!) Any comments by WRI well informative persons (and other of course!) will be greatly appreciate.
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- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
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