Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

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.



  • Prev by Date: Audio
  • Next by Date: Re: 6.0 Get Graphics Coordinates...
  • Previous by thread: Re: Audio
  • Next by thread: Re: Integrate modified in version 6?