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Re: Double integral of a piecewise-constant function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg61556] Re: Double integral of a piecewise-constant function
  • From: Bhuvanesh <lalu_bhatt at yahoo.com>
  • Date: Sat, 22 Oct 2005 00:36:02 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

A single Integrate command should work.

expr = Piecewise[{{0, t[2] < 0 || t[2] >= 3 || (t[2] < 1 
&& t[1] < 1) || (t[2] < 2 && t[1] >= 3) || (t[2] >= 1 && t[1] < 0) ||
(Inequality[1, LessEqual, t[2], Less, 2] && Inequality[1, 
LessEqual, t[1], Less, 2]) || (t[2] >= 2 && t[1] >= 2)},
{Comm[Subscript[H, 1], Subscript[H, 2]], t[2] < 1 && t[1] < 2}, 
{Comm[Subscript[H, 1], Subscript[H, 3]], t[2] < 1},
{Comm[Subscript[H, 2], Subscript[H, 1]], t[2] < 2 && t[1] < 1}, 
{Comm[Subscript[H, 2], Subscript[H, 3]], t[2] < 2},
{Comm[Subscript[H, 3], Subscript[H, 1]], t[1] < 1}}, 
Comm[Subscript[H, 3], Subscript[H, 2]]];

-Integrate[expr, {t[1], 0, 3}, {t[2], 0, t[1]}]/2

Bhuvanesh,
Wolfram Research.


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