Re: Integrate[], Sort[] and Hold[]

*To*: mathgroup at smc.vnet.net*Subject*: [mg101012] Re: [mg100994] Integrate[], Sort[] and Hold[]*From*: "David Park" <djmpark at comcast.net>*Date*: Sat, 20 Jun 2009 04:03:04 -0400 (EDT)*References*: <24964246.1245461320010.JavaMail.root@n11>

I was somewhat surprised that the following worked: Integrate[Min[2^a, 3^a], {a, -1, 1}] % // N (2 Log[2] + Log[27])/(Log[2] Log[27]) 2.04952 NIntegrate[Min[2^a, 3^a], {a, -1, 1}] 2.04952 David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ From: Neil Stewart [mailto:neil.stewart at warwick.ac.uk] I am having trouble integrating a function which contains Sort[]. Here is a stripped down (if slightly odd) example: In = Integrate[Sort[{2, 3}^a], {a, -1, 1}] Out = {3/Log[4], 8/Log[27]} Mathematica is evaluating this as follows. First, Sort[{2, 3}^a] is evaluated as {2^a, 3^a}. Then Integrate[2^a, {a, -1, 1}] gives the first term 3/Log[4]. Finally Integrate[3^a, {a, -1, 1}] gives the second term 8/Log[27]. Note here that Sort[] is sorting 2^a and 3^a without knowing the value of a. That is, sort is sorting the raw symbolic expressions. I would prefer Sort[] to wait until it knows the value of a before sorting. For example, when a is -1, then 2^a = 1/2 and 3^a = 1/3, so Sort[{2, 3}^a] would be {1/3, 1/2}. However when a is 1, then 2^a = 2 and 3^a = 3, so Sort[{2, 3}^a] would be {2, 3}. That is, in the first case the terms are swapped, but in the second case they are not. So what I'm after is In = Integrate[Sort[{2, 3}^a], {a, -1, 1}] Out = {2/Log[3] + 1/Log[4], 1/Log[2] + 2/Log[27]} (* This does not actually happen *) [If you prefer to picture this, Plot[{2^a, 3^a}, {a, -1, 1}] draws two increasing lines that cross at a = 0. Mathematica is integrating under each curve. I'm trying to integrate under the line made from the two lower segments, and under the line made from the two upper segments.] I've tried using Hold[], ReleaseHold[], and Evaluate[] but have got myself into a terrible mess. Obviously with this trivial example I could just split the integral up myself, but is there a way to achieve delaying Sort[] until a is known? I would be very grateful for any comments.

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**Re: Integrate[], Sort[] and Hold[]**

**Re: Integrate[], Sort[] and Hold[]**