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MathGroup Archive 2007

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Re: Integrate modified in version 6?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77976] Re: [mg77938] Integrate modified in version 6?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 21 Jun 2007 05:28:25 -0400 (EDT)
  • References: <200706200930.FAA09746@smc.vnet.net>

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.
>
>



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