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Re: evaluate assuming odd integers?


Simplify[Integrate[Sin[(2m+1)*Pi*(z/L)], {z, 0, 
L}],Assumptions->m\[Element]Integers]

(2*L)/(Pi + 2*m*Pi)

Steve Luttrell

"ab at sd.com" <at2 at ads.com> wrote in message news:dc1t2f$4v8$1 at smc.vnet.net...
> Is there a way in Mathematica to evaluate an expression assuming some
> varialbe is Odd integer, for example in below integral how can i tell it
> that n is odd integer in the Assumptions section of FullSimplify command 
> or
> the integral itself??:
> In[35]:=
> Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
> FullSimplify[%, n \[Element] Integers && n >= 0 &&
> L \[Element] Reals && L > 0 && z \[Element] Reals]
> 



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