Re: wrong solution for double integral of piecewise function?
- To: mathgroup at smc.vnet.net
- Subject: [mg96993] Re: wrong solution for double integral of piecewise function?
- From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
- Date: Sun, 1 Mar 2009 04:56:45 -0500 (EST)
- References: <gob80u$g9q$1@smc.vnet.net>
Tom,
Why didn't you look up UnitStep? One press on F1 (PC) and you're done.
It shows you that the answers are the same if a<b. I guess you have
made the same assumption in your calculation.
Cheers -- Sjoerd
On Feb 28, 1:43 pm, Tom Roche <tlro... at gmail.com> wrote:
> 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
>
> Solve[
> First[
> Integrate[
> Integrate[
> 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?