Re: Why is Mathematica assuming k==l and how do I make it not to?

*To*: mathgroup at smc.vnet.net*Subject*: [mg92612] Re: Why is Mathematica assuming k==l and how do I make it not to?*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Wed, 8 Oct 2008 06:25:23 -0400 (EDT)*References*: <gcffpl$f4g$1@smc.vnet.net>

Aaron Fude <aaronfude at gmail.com> wrote: > As in > > Assuming[Element[{k, l}, Integers] , > Integrate[Cos[k alpha] Cos[l alpha], {alpha, -Pi, Pi}]] > > I get 0 whereas the answer is Pi if k=l; Your sort of question arises fairly often in this newsgroup. For example, last month, it arose in the thread "Integration in Mathematica". See my response then at <http://groups.google.com/group/comp.soft-sys.math.mathematica/msg/1ff92f03d99d99e9>. In your specific example, it seems that Mathematica is assuming that k and l are different, even though they could be equal. Thus, I suppose you really intended your title to be "Why is Mathematica assuming k != l and how do I make it not to?" The basic trouble in your example lies not in k and l being integers or in the definite integral, but rather in the indefinite integral In[3]:= Integrate[Cos[k alpha] Cos[l alpha], alpha] Out[3]= Sin[alpha*(k - l)]/(2*(k - l)) + Sin[alpha*(k + l)]/(2*(k + l)) Note that the above is problematic if |k| = |l|. This problem could have been avoided if Mathematica had given the result in terms of the sine cardinal function, Sinc, newly implemented in version 6. Specifically, the indefinite integral is given by indef[alpha_, k_, l_] := alpha/2*(Sinc[alpha (k-l)] + Sinc[alpha (k+l)]); Then, using the Fundamental Theorem, your desired definite integral is In[6]:= def = indef[Pi, k, l] - indef[-Pi, k, l] Out[6]= Pi*(Sinc[(k - l)*Pi] + Sinc[(k + l)*Pi]) If k and l are integers, then the above yields 0 or Pi, depending resp. on whether the absolute values of k and l are different or the same. David W. Cantrell