Re: Integrate modified in version 6?

• To: mathgroup at smc.vnet.net
• Subject: [mg78113] Re: Integrate modified in version 6?
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Sat, 23 Jun 2007 07:08:11 -0400 (EDT)
• References: <200706200930.FAA09746@smc.vnet.net> <f5dgdj\$nlh\$1@smc.vnet.net>

```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.

How is that clear?

Also note that Bob H. and Curtis O. both report getting -Infinity as the
result in version 6 (which I don't have).

FWIW, using version 5.1, FullSimplify applied to those two MeijerG terms
yields Indeterminate and ComplexInfinity, resp.

My guess is that the output from version 6 involving MeijerG might be pure
nonsense. If so, maybe that's why it takes "for ever" to evaluate
numerically.

In any event, version 5.1 gets a nice answer easily:

In[1]:= Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
Out[1]= -1 - 2*Catalan + Pi*(1/4 + Log[2])

In[2]:= N[%]
Out[2]= 0.131053

In[3]:= NIntegrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
Out[3]= 0.131053

and so the behavior of version 6 on this particular integral is clearly not
as good.

David W. Cantrell

```

• Prev by Date: Re: save as pdf in version 6
• Next by Date: Re: Combination List
• Previous by thread: Re: Re: Integrate modified in version 6?
• Next by thread: Re: Re: Integrate modified in version 6?