Re: Assuming and Integrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg128362] Re: Assuming and Integrate*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Tue, 9 Oct 2012 00:41:08 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <20121006055312.C0D5B6908@smc.vnet.net> <k4r4en$noo$1@smc.vnet.net> <20121008062943.DD0416837@smc.vnet.net>

On Oct 8, 2012, at 2:29 AM, Roland Franzius <roland.franzius at uos.de> wrote: > Am 07.10.2012 07:37, schrieb Murray Eisenberg: >> How to treat exceptional cases, here m = n an integer, is always an issue with Integrate (and other operations). Here one would hope to see at least a ConditionalExpression for the general situation Element[{n,m}, Integers]. After all, such integrals are so common in Fourier analysis. >> >> Perhaps this should even be reported as a bug to Wolfram Research. >> >> On Oct 6, 2012, at 1:53 AM, hamiltoncycle at gmail.com wrote: >> >>> When I try the line below in Mathematica 8 I get the answer 0 which is what I expect when m and n are different but not when m=n. Can anyone explain how to do this correctly? >>> >>> Assuming[Element[{n, m}, Integers], Integrate[Sin[n*x]*Sin[m*x], {x, 0, Pi}]] >>> >>> When m=n we should get Pi/2, as in this case: >>> >>> Assuming[Element[n, Integers], Integrate[Sin[n*x]*Sin[n*x], {x, 0, Pi}]] >>> > > > > For integration you have to use the specialized Assumptions option of > Integrate. > > Integrate[Sin[n x] Sin[m x],{x,0,Pi},Assumptions->{n,m} \in Integers] > is giving the correct Kronecker symbol 0 or +-pi/2 for for n=+-m. > > The reason will be that - with the exception of linear real > substitutions - external assumptions cannot be translated to domain > conditions in substitutions in the complex domains or through > transcendental replacements of variables. > > Instead the integration machine has to plan ist table lookup or > simplification and transformation process with the special domain > assumptions to be designed by hand for an successful attempt. > > > The half period sines are a complete basis of the Hilbert space on > (0,Pi) but in Fourier analysis they are used only for zero boundary > conditions. So the orthogonality relations are not so evident contrary > to the every day formulas for standard periodic boundary conditions. > I don't know which version of Mathematica you're using, but in 8.0.4, one does NOT get any kind of conditional (hence correct) result from: Integrate[Sin[n x] Sin[m x],{x,0,Pi},Assumptions->{n,m}\[Element] Integers] (Note the syntax correction from "\in" to "\[Element]" there.) Rather, one gets the same "general" result as from the OP's Assuming expression, namely: (n Cos[n \[Pi]] Sin[m \[Pi]]-m Cos[m \[Pi]] Sin[n \[Pi]])/(m^2-n^2) And as originally said, this is wrong when m = n. --- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305

**References**:**Assuming and Integrate***From:*hamiltoncycle@gmail.com

**Re: Assuming and Integrate***From:*Roland Franzius <roland.franzius@uos.de>