Re: Integrate modified in version 6?
- To: mathgroup at smc.vnet.net
- Subject: [mg78110] Re: Integrate modified in version 6?
- From: dimitris <dimmechan at yahoo.com>
- Date: Sat, 23 Jun 2007 07:06:39 -0400 (EDT)
- References: <f5asi1$9t6$1@smc.vnet.net>
So here it seems:
We have a platform dependent integral.
However, my original query
was if the integrate algorithm
has been modified from 5.2 to 6.
The adding rule for Limit shows
that in version 5.2 NL formula is
applied to the resulting antiderivative.
In version 6 the integral seems to be
evaluated through Slater convolution.
Dimitris
dimitris :
> 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.