Re: A simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg47618] Re: A simple integral
- From: drbob at bigfoot.com (Bobby R. Treat)
- Date: Sun, 18 Apr 2004 04:15:15 -0400 (EDT)
- References: <c5qj2s$fvc$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
The negative sign is easy to explain, since it appears under the
assumption that mu is negative -- hence the result is positive WITH
that negative sign.
Version 5.0.1 gives yet another answer:
Integrate[x^2/E^((x - \[Micro])^2/
(2*s^2)), {x, -8, 8}]
(1/2)*s*((2*s*(-8 + \[Micro]))/
E^((8 + \[Micro])^2/(2*s^2)) -
(2*s*(8 + \[Micro]))/
E^((-8 + \[Micro])^2/(2*s^2)) -
Sqrt[2*Pi]*(s^2 + \[Micro]^2)*
Erf[(-8 + \[Micro])/(Sqrt[2]*
s)] + Sqrt[2*Pi]*
(s^2 + \[Micro]^2)*
Erf[(8 + \[Micro])/(Sqrt[2]*s)])
Did you mean Infinity when you posted 8 in that integral? If so,
Integrate[x^2/E^((x - \[Micro])^2/
(2*s^2)), {x, -Infinity,
Infinity}]
If[Re[s^2] > 0 && Re[\[Micro]/s^2] <
0, 0, Integrate[
x^2/E^((x - \[Micro])^2/(2*s^2)),
{x, -Infinity, Infinity},
Assumptions ->
Re[s^2] <= 0 ||
Re[\[Micro]/s^2] >= 0]]
The second argument of THAT answer is manifestly wrong, since the
integral can't be zero. (And the third argument seems rather useless.)
Yes, v5 has apparently caused more problems than it solved.
Bobby
"Dr A.H. Harker" <a.harker at ucl.ac.uk> wrote in message news:<c5qj2s$fvc$1 at smc.vnet.net>...
> A simple integration, under Version 4.1.2:
>
> Integrate[x^2 Exp[-(x-$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