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wrong solution for double integral of piecewise function?

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  • Subject: [mg96968] wrong solution for double integral of piecewise function?
  • From: Tom Roche <tlroche at>
  • Date: Sat, 28 Feb 2009 06:43:39 -0500 (EST)

wrong solution for double integral of piecewise function

I've got a function

f[\[Chi]_, \[Psi]_] =
  {Piecewise[{{k, {a <= \[Chi] <= b, a <= \[Psi] <= b}}}, 0]}

I'm attempting to solve for k such that f becomes a probability
density function by applying the normalization constraint, i.e.
solving for k such that the double (indefinite) integral of f equals
1. I can do this by hand pretty easily, integrating first/inside WRT
psi and second/outside WRT chi, and I get

(1) k = 1/((b-a)^2)

I don't have the world's greatest calculus chops, but that looks
correct to me. (Am I missing something?) However, when I use
Mathematica to

        f[\[Chi], \[Psi]], {\[Psi], -\[Infinity], \[Infinity]}
        {\[Chi], -\[Infinity], \[Infinity]}
  ] == 1, k

I get

(2) {{k -> -(1/((a - b)^2 (-1 + UnitStep[a - b])))}}

which seems wrong to me, though I'll admit I don't know what
"UnitStep" means. So I'm wondering

* does (1) = (2)? or

* have I made a syntax error? or

* is this just really hard to solve symbolically? If so, is there a
  better way to setup this function and its solution?

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