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MathGroup Archive 2002

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RE: Re: Yet another Version 4.2 Integration howler

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35318] RE: [mg35301] Re: Yet another Version 4.2 Integration howler
  • From: "DrBob" <majort at cox-internet.com>
  • Date: Mon, 8 Jul 2002 03:16:39 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

Not to worry:

$Version

4.2 for Microsoft Windows (June 5, 2002)

Integrate[Log[z]/(1 + z^2), {z, 0,1}]

-Catalan

Integrate[Log[z]/(1 + z^2), {z, 1,Infinity}]

Catalan

Integrate[Log[z]/(1 + z^2), {z, 0,Infinity}]

0

Bobby Treat

-----Original Message-----
From: Kevin J. McCann [mailto:kjm at KevinMcCann.com] 
To: mathgroup at smc.vnet.net
Subject: [mg35318] [mg35301] Re: Yet another Version 4.2 Integration howler

Vladimir Bondarenko wrote:

> 
........................................................................
.......
> 
> BUG # 3617
> 
> For an expression having a concrete numeric value,  Indeterminate is
> reported
> 
> BUG HISTORY:    PRESENT   4.2 for Microsoft Windows (February 28,
2002)
>                 ABSENT    4.1 for Microsoft Windows (November 2, 2000)
>                 ABSENT    4.0 for Microsoft Windows (April 21, 1999)
>                 ABSENT    Microsoft Windows 3.0 (April 25, 1997)
>                 ABSENT    Windows 387 2.2 (April 9, 1993)
> 
> Reproducible    Always
> BuildNumber     156656
> CPU ID          AuthenticAMD AMD Athlon(tm) XP 1600+
> RAM size        512 Mb
> Free HDD size   11  Gb
> OS ID           Microsoft Windows 98 4.10.2222 A

In fact, the integration from 1 to infinity is Catalan's constant, and,
of 
course, the integral from 0 to 1 of the integrand is the negative of it.

Nice upgrade ;(

Kevin

> 
> The value of the integral is defined.
> However, Mathematica reports  Indeterminate  which is invalid.
> 
>                 Integrate[Log[z]/(1 + z^2), {z, 0, Infinity}]
> 
> ACTUAL:         Indeterminate
> 
> EXPECTED:       0
> 
> CHECK-UP:       NIntegrate[Log[z]/(1 + z^2), {z, 0, Infinity}] // Chop
> 
>                 0
> 
> 
........................................................................
.......
> 
> 
> Best,
> 
> Vladimir Bondarenko





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