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Re: Correct answer for all n

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  • Subject: [mg113231] Re: Correct answer for all n
  • From: Daniel Lichtblau <danl at>
  • Date: Wed, 20 Oct 2010 04:06:04 -0400 (EDT)

Sam Takoy wrote:
> Hi,
> Assuming[Element[n, Integers],
>   1/(2 \[Pi]) Integrate[Sin[a]*Exp[-I*n*a], {a, -Pi, Pi}]]
> returns 0, where as for n = 1 the answer is not zero. Anyway to have 
> Mathematica acknowledge that automatically?
> Many thanks in advance,
> Sam

This came up earlier in the month.


(1) Integrality assumptions will invoke simplifiers that do not 
recognize special cases (which can be regarded, in some sense, as 

(2) If you forego that assumption, you can recover correct results for 
specific integer n using Limit. Example:

In[3]:= InputForm[ii =
   1/(2*Pi)*Integrate[Sin[a]*Exp[-I*n*a], {a,-Pi,Pi}]]
Out[3]//InputForm= (I*Sin[n*Pi])/((-1 + n^2)*Pi)

In[5]:= Limit[ii,n->0]
Out[5]= 0

In[7]:= InputForm[Limit[ii,n->-1]]
Out[7]//InputForm= I/2

Daniel Lichtblau
Wolfram Research

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