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MathGroup Archive 1999

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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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