Re: Re: Integrate modified in version 6?
- To: mathgroup at smc.vnet.net
- Subject: [mg78096] Re: [mg77976] Re: [mg77938] Integrate modified in version 6?
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 22 Jun 2007 06:53:41 -0400 (EDT)
- Reply-to: hanlonr at cox.net
I get -Infinity on my machine with v6. I never had a copy of the beta version.
$Version
6.0 for Mac OS X x86 (32-bit) (April 20, 2007)
Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
-Infinity
Bob Hanlon
---- Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> My version of Mathematica 6.0 gives:
>
> Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
>
> (1/2)*(MeijerG[{{-(1/2), 0}, {1}}, {{0, 0, 1/2}, {}}, 1] -
> MeijerG[{{-(1/2), 0}, {1, 1}}, {{0, 0, 1/2, 1/2}, {}}, 1]/Sqrt
> [Pi])
>
> I don't know whether this answer is right or wrong (Mathematica takes
> for ever to evaluate this numerically) but one thing is clear: it
> certainly is not -Infinity. Which only proves, what you would have
> already known if you read some past posts on similar topics, namely,
> that you can't trust the person you got this information from because
> he has a habit of using beta versions long past the release date and
> then calling bugs he finds in the beta version as "bugs in
> Mathematica". He has done this before (on this forum) and it seems he
> is still doing it on others.
>
> Andrzej Kozlowski
>
>
> On 20 Jun 2007, at 18:30, dimitris wrote:
>
> > 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.
> >
> >
>
>