       Re: Bug of Integrate

• To: mathgroup at smc.vnet.net
• Subject: [mg83019] Re: Bug of Integrate
• From: Miguel <misvrne at gmail.com>
• Date: Wed, 7 Nov 2007 06:53:48 -0500 (EST)
• References: <fg4dfv\$6c3\$1@smc.vnet.net><fg6pse\$d44\$1@smc.vnet.net>

```On 1 nov, 11:38, "David W.Cantrell" <DWCantr... at sigmaxi.net> wrote:
> m... at inbox.ru wrote:
> > On Oct 30, 2:26 am, "David W.Cantrell" <DWCantr... at sigmaxi.net> wrote:
> [snip]
> > > However, related to the above, version 5.2 does give an incorrect
> > > result for a definite integral with a symbolic real limit. Whether this
> > > error still exists in version 6, I don't know:
>
> > > In:= Assuming[Element[x,Reals],Integrate[3*Sign[Cos[t]],{t,0,x}]]
>
> > > Out= 3 If[x > 0, x Abs[Cos[x]] Sec[x],
> > > Integrate[Sign[Cos[t]], {t, 0, x}, Assumptions -> x <= 0]]
>
> > > The above is incorrect for x > Pi/2. A correct result would have been
>
> > > 3 Sign[Cos[x]] (x - Pi Floor[x/Pi + 1/2])
>
> > > for all real x.
>
> > > David W. Cantrell
>
> > Note that your formula isn't correct for x = Pi/2 + Pi k. The correct
> > expression for all real x is
>
> > In:= Assuming[0 <= x < 2 Pi, Integrate[3 Sign[Cos[t]], {t, 0,
> > x}]] /.
> >   x -> Mod[x, 2 Pi]
>
> > Out= Piecewise[{{-3 Pi/2, Mod[x, 2 Pi] == 3 Pi/2}, {3 (Pi - Mod[x,
> > 2 Pi]), Pi/2 < Mod[x, 2 Pi] < 3 Pi/2}, {-3 (2 Pi - Mod[x, 2 Pi]), 3 Pi/
> > 2 < Mod[x, 2 Pi] < 2 Pi}, {3 Mod[x, 2 Pi], 0 < Mod[x, 2 Pi] <= Pi/2}}]
>
> Moments ago, I sent a message thanking Maxim for pointing out my error. I
> also mentioned a much shorter result which is correct for all real x:
>
> 3 ArcSin[Sin[x]]
>
> But perhaps it's worth mentioning that there is also an expression
> which is correct for all real x which avoids using any functions such as
> ArcSin or Sin, while still being shorter than his Piecewise expression:
>
> 3 (-1)^Floor[x/Pi + 1/2] (x - Pi Floor[x/Pi + 1/2])
>
> That result is "in the same spirit" as what I originally intended.
>
> David W. Cantrell- Ocultar texto de la cita -
>
> - Mostrar texto de la cita -

By other hand, I think everybody knows the solution of the following
problem:
"Derive the formula for the circunference of a circle of radius "a" by
computing the length of the arc     x=a cost; y= a sint  for 0<=t<=Pi
"

L=Integrate[Sqrt[1+(y'[t]/x'[t])^2]*x'[t],{t,0,2Pi}]

I have tried to resolve with version 5.2 and version 6.0.1. The
results have been differents.

When the result is unknown, version 6.0.1 is reliable?

```

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