Re: Help -- Weird integration behavior
- To: mathgroup at smc.vnet.net
- Subject: [mg49358] Re: Help -- Weird integration behavior
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Fri, 16 Jul 2004 06:06:39 -0400 (EDT)
- References: <cd5p22$b5c$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
miked378 at hotmail.com (Mike) wrote:
> I'm working on a project that I last worked on a couple of years ago,
> and I'm trying to pick up where I left off. I had been doing some
> integrations using Mathematica 4.2, and all was well. I'm now trying
> to use the same workbook, but with Mathematica 5.0, and the
> integration that previously seemed to work now gives me a value of
> zero.
>
> Interestingly, when I run the same file on different versions of
> Mathematica, I get either zero or an expression:
>
> --v5.0 (student), Macintosh: zero
> --v5.0, Solaris: zero
> --v4.2, Macintosh: expression
> --v4.0, Windows: expression
>
> Here's the code:
> F[x_, t_] = (ks (UnitStep[x] - UnitStep[x - 0.5]))(UnitStep[t] -
> UnitStep[t - 0.5]);
> phi[n_, x_] = Sqrt[2] Sin[n Pi x/L]
> G[n_, t_] = Integrate[F[x, t] phi[n, x], {x, 0, L}]
>
> It's the output of this last line that's interesting to me -- either
> an expression (I hope) or zero. My questions are:
>
> -- Did something change in 5.0 that makes it so that this integration
> doesn't work correctly?
So it would seem.
> -- Can anybody out there reproduce my problem (i.e., run these three
> cells on different versions and produce different answers?)
Well, running it in v5.0.0 (Windows), I get 0 .
Let me guess that your L is always greater than 0. If that guess is
correct, then notice the following in v5.0.0, adding an assumption (and
replacing your 0.5's by 1/2's):
In[1]:=
F[x_, t_] = (ks (UnitStep[x] - UnitStep[x - 1/2]))(UnitStep[t] -
UnitStep[t - 1/2]);
phi[n_, x_] = Sqrt[2] Sin[n Pi x/L];
G[n_, t_] = Assuming[L>0,Integrate[F[x, t] phi[n, x], {x, 0,L}]]
Out[1]=
-((1/(n*Pi))*(Sqrt[2]*ks*L*(1 - Cos[n*Pi] + (Cos[n*Pi] -
Cos[(n*Pi)/(2*L)])* UnitStep[-(1/2) + L])*(UnitStep[-(1/2) + t] -
UnitStep[t])))
I have not checked to see if that result is correct.
HTH,
David Cantrell