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Re: Multi-variable Integration

• To: mathgroup at smc.vnet.net
• Subject: [mg79711] Re: Multi-variable Integration
• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
• Date: Thu, 2 Aug 2007 03:54:27 -0400 (EDT)
• Organization: Uni Leipzig
• References: <f8pjji$2ae$1@smc.vnet.net>
• Reply-to: kuska at informatik.uni-leipzig.de

Hi,

form the mathematical point of view the general convention
is that the integration constant for indefinite integration
is zero ... otherwise

Integrate[D[a*b*c*d*e*x*y,x],x]

would be

a*b*c*d*e*x*y+arbitaryConstant[a,b,c,d,e,y]

clearly this would made the system unusable because it would
quickly grow the complexity of the expression.

In the dedicated cases, where you wish to violate the
convention about the integration constant, you must do
it by hand.

Regards
   Jens


gravmath at yahoo.com wrote:
> Suppose one defines two expressions in Mathematica:
> 
> Q = f[x,y]
> P = f[x,y] + g[y]
> 
> and then differentiates them wrt the variable 'x' as follows:
> 
> dQ = D[Q,x]
> dP = D[P,x]
> 
> Subsequent use of the Integrate command wrt the variable
> 'x' (Integrate[dQ,x] and Integrate[dP,x]) yields, in both cases,
> f[x,y].
> 
> I find this behavior understandable from a systems point of view but
> mathematically in both cases the answer should be f[x,y] +
> arbitraryfunc[y], where obviously further input (as in the original
> definitions of P & Q) is needed to determine arbitraryfunc[y].  Is
> there a way to get Mathematica to recognize that there are two
> variables in the problem and to produce the arbitrary function of the
> variable 'y'?
> 
> I'm guessing that my specification of f[x,y] is not quite sufficient
> to do this, even though it is sufficient when differentiating.  That
> is to say that dQ and dP are rendered in Mathematica as f^(1,0)[x,y],
> which clearly indicates that Mathematica understands that there are
> two independent variables in the expression.
> 
> Any help would be appreciated, even if it to point me to previous
> posts (I found no germane ones myself).
> 
> Thanks in advance,
> Conrad
> 
>