Re: Correct answer for all n

*To*: mathgroup at smc.vnet.net*Subject*: [mg113231] Re: Correct answer for all n*From*: Daniel Lichtblau <danl at wolfram.com>*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. http://forums.wolfram.com/mathgroup/archive/2010/Oct/msg00119.html Recap: (1) Integrality assumptions will invoke simplifiers that do not recognize special cases (which can be regarded, in some sense, as "nongeneric"). (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