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Re: Assumptions in Integrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg20584] Re: [mg20562] Assumptions in Integrate
*From*: BobHanlon at aol.com
*Date*: Sat, 30 Oct 1999 14:54:57 -0400
*Sender*: owner-wri-mathgroup at wolfram.com
Julian,
Integrate[Cos[n x], {x, -Pi, Pi}, Assumptions -> Element[n, Integers]]
(2*Sin[Sqrt[n^2]*Pi])/Sqrt[n^2]
Integrate[Cos[n x], {x, -Pi, Pi}, Assumptions -> n >= 0]
(2*Sin[n*Pi])/n
To get Mathematica to "evaluate" the Sqrt, it must believe that n is
nonnegative
Integrate[Cos[n x], {x, -Pi, Pi},
Assumptions -> {Element[n, Integers] , n >= 0}]
(2*Sin[n*Pi])/n
Integrate[Cos[n x], {x, -Pi, Pi},
Assumptions -> {Element[n, Integers] , n > 0}]
(2*Sin[n*Pi])/n
Bob Hanlon
In a message dated 10/30/1999 7:14:07 AM, mtpagesj at lg.ehu.es writes:
> I find the results of using assumptions in Integrate somewhat
>stranege. For instance,
>
>In[21]:= Integrate[Cos[n x], {x, -Pi, Pi}]
>
>Out[21]= 2 Sin[n Pi]
> -----------
> n
>
>In[22]:= Integrate[Cos[n x], {x, -Pi, Pi},
> Assumptions -> Element[n, Integers]]
>
>Out[22]=
> 2
>2 Sin[Sqrt[n ] Pi]
>------------------
> 2
> Sqrt[n ]
>
>In[23]:= $Version
>
>Out[23]= "4.0 for Power Macintosh (July 20, 1999)"
>
> I know I can define my own transformation rules, but one would
>think that Mathematica should do it directly.
>
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