Re: Assuming and Integrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg128335] Re: Assuming and Integrate*From*: Roland Franzius <roland.franzius at uos.de>*Date*: Mon, 8 Oct 2012 02:29:43 -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>

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. -- Roland Franzius

**Follow-Ups**:**Re: Assuming and Integrate***From:*Murray Eisenberg <murray@math.umass.edu>

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