Re: A Problem with Simplify

• To: mathgroup at smc.vnet.net
• Subject: [mg87687] Re: A Problem with Simplify
• From: Alexey Popkov <popkov at gmail.com>
• Date: Tue, 15 Apr 2008 06:49:35 -0400 (EDT)
• References: <fu1u56\$onj\$1@smc.vnet.net>

```On 15 =C1=D0=D2, 13:56, Bill Rowe <readn... at sbcglobal.net> wrote:
> On 4/14/08 at 5:43 AM, pop... at gmail.com (Alexey Popkov) wrote:
>
> >Try the following:
> >Integrate[Exp[(a - 1)*x], x] /. a -> 1
> >Integrate[Cos[(a - 1)*x], x] /. a -> 1
> >Integrate[(a - 1)^x, {x, -1, 0}] /. a -> 1
> >Integrate[Cos[a x]/Sin[x], x] /. a -> 1
> >There is the ONE underlying BUG!
>
> There is no bug here. Each of the integrals you do above result
> in an expression divided by (a-1). The replacement rule simply
> substitutes 1 for a everywhere. Consequently, the denominator of
> each integral evaluates to 0 and Mathematica returns either
> Indeterminate or ComplexInfinity as it should.
>
> Where you expecting a to be replaced by 1 before the integration
> was done? If so, the syntax should be written
>
> Integrate[f/.a->1,x]
>
> Or perhaps you were expecting Mathematica to simplify the
> results of the integral before replacing a with 1. If so, you
> need either
>
> FullSimplify[Integrate[f,x]]/.a->1
>
> or
>
> Simplify[Integrate[f,x]]/.a->1
>
> However, neither of these are guaranteed to remove the
> singularity. So, these may result in exactly what you got.

I am expecting to get all partial answers or at least an answer with
conditions!

On 11 =C1=D0=D2, 09:42, dh <d... at metrohm.ch> wrote:
>In the manual one finds:
> "For indefinite integrals, Integrate tries to find results that are
> correct for almost all values of parameters."
for Integrate. And this seems to be false... :( And for definite
integrals there is an option "GenerateConditions" that does not work
in the most cases (try it with example above:
int1 = Integrate[Sin[a*x/L]*Sin[x/L], {x, 0, L}, GenerateConditions ->
True]
int1 /. a -> 1
The answer is Indeterminate that is false! :(
).

```

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