A beginer's simple question about Mathematica...
- To: mathgroup at smc.vnet.net
- Subject: [mg63540] A beginer's simple question about Mathematica...
- From: leelsuc at yahoo.com
- Date: Wed, 4 Jan 2006 03:17:19 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
I try to run
Integrate[x*Exp[-((x-2*a)^2/(8*a)+(y-b)^2/(2*b)+(z*x-y-c)^2/(2*c))],{x,-\[infinity],0},{y,-\[infinity],+\[infinity]}]
get
\!\(\(\(1\/\@\(1\/b +
1\/c\)\)\((\@\(2\ p\)\ If[Re[z] > \(-\(1\/2\)\) && Re[\(b + c +
4\ a\ z\
\^2\)\/\(8\ a\ b +
8\ a\ c\)] >
0, \(-\(\(2\ \[ExponentialE]\^\(1\/2\ \((\(-a\) - b -
\
c)\)\)\ \((\(-\[ExponentialE]\^\(\(a\ \((b + c)\)\ \((1 + 2\
z)\)\^2\)\/\(2\ \
\((b + c + 4\
a\ z\^2)\)\)\)\)\ \@\(2\ p\)\ \((1 + 2\ z)\) + 2\
\@\(\(b + c \
+ 4\ a\ z\^2\)\/\(a\ b + a\ c\)\) + \[ExponentialE]\^\(\(a\ \((b +
c)\)\ \((1 \
+ 2\ z)\)\^2\)\/\(2\ \((b + c + 4\ a\ z\^2)\)\)\)\ \@\(2\ p\)\ \((1 +
2\ z)\)\
\ Erf[\(1 + 2\ z\)\/\(\@2\ \@\(\(b + c + 4\ a\ z\^2\)\/\(a\ b + a\ \
c\)\)\)])\)\)\/\((\(b + c + 4\ a\ z\^2\)\/\(a\ b + a\
c\))\)\^\(3/2\)\)\),
Integrate[\(-\[ExponentialE]\^\(-\(\(4\ a\^2\ \((b
+ c)\) \
+ \((b + c)\)\ x\^2 + 4\ a\ \((b\^2 + c\^2 +
x\^2\ z\^2 - c\ \((\(-x\) - 2\ x\ z)\) + b\ \((2\ c
+ x\ \
\((1 + 2\ z)\))\))\)\)\/\(8\ a\ \((b + c)\)\)\)\)\)\ x, {x, 0, 8}, \
Assumptions -> Re[z] = \(-\(1\/2\)\) || \((Re[z] > \(-\(1\/2\)\) &&
Re[\(b + \
c + 4\ a\ z\^2\)\/\(8\ a\ b + 8\ a\ c\)] = 0)\)]])\)\)\)
What does this result mean?
Thanks a lot for replying.