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Re: A simple integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47606] Re: [mg47596] A simple integral
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sun, 18 Apr 2004 04:15:03 -0400 (EDT)
  • References: <200404170631.CAA16293@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 17 Apr 2004, at 15:31, Dr A.H. Harker wrote:

> A simple integration, under Version 4.1.2:
>
> Integrate[x^2 Exp[-(x-$B&L(B)^2/(2 $B&R(B^2)],{x,-$B!g(B,$B!g(B}]
>
>        2
> If[Re[$B&R(B ] > 0,
>
>                    2
>   Sqrt[2 Pi] Sqrt[$B&R(B ]
>
>      2    2
>    ($B&L(B  + $B&R(B ),
>
>                     2
>                    x
>   Integrate[----------------,
>                     2     2
>              (x - $B&L(B) /(2 $B&R(B )
>             E
>
>    {x, -Infinity, Infinity}]]
>
> and the same under 5.0
>
> Integrate[x^2 Exp[-(x-$B&L(B)^2/(2 $B&R(B^2)],{x,-$B!g(B,$B!g(B}]
>
>        2            $B&L(B
> If[Re[$B&R(B ] > 0 && Re[--] < 0,
>                      2
>                     $B&R(B
>
>                    2    2
>     Sqrt[2 Pi] $B&L(B ($B&L(B  + $B&R(B )
>   -(----------------------),
>                  2
>                 $B&L(B
>            Sqrt[--]
>                  2
>                 $B&R(B
>
>                     2
>                    x
>   Integrate[----------------,
>                     2     2
>              (x - $B&L(B) /(2 $B&R(B )
>             E
>
>    {x, -Infinity, Infinity},
>
>    Assumptions ->
>
>        $B&L(B               2
>     Re[--] >= 0 || Re[$B&R(B ] <= 0
>         2
>        $B&R(B
>
>     ]]
>
> Two questions:
>     1. Whence the extra condition in Version 5?
>     2. Why the negative sign in Version 5? Using PowerExpand then gives
> a negative result for this integral which is patently, for real
> parameters, positive.
>
> Am I alone in feeling that Version 5 has introduced more problems than
> it has solved?
>
>  Dr A.H. Harker
>  Department of Physics and Astronomy
>  University College London
>  Gower Street
>  LONDON
>  WC1E  6BT
>  (44)(0)207 679 3404
>  a.harker at ucl.ac.uk
>
>
The price of progress ;-)

However, note that (in 5.0)


Integrate[x^2*Exp[-(x - $B&L(B)^2/(2*$B&R(B^2)], {x, -Infinity, Infinity}, 
Assumptions -> {$B&R(B > 0}]


Sqrt[2*Pi]*$B&R(B*($B&L(B^2 + $B&R(B^2)



Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/


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