Re: Assuming and Integrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg128350] Re: Assuming and Integrate*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Tue, 9 Oct 2012 00:37:07 -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>

On 8 Okt., 08:52, Roland Franzius <roland.franz... 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, hamiltoncy... 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 I get in version 8: In[3]:= Integrate[Sin[n*x]*Sin[m*x], {x, 0, Pi}, Assumptions -> Element[{n, m}, Integers]] Out[3]= (n*Cos[n*Pi]*Sin[m*Pi] - m*Cos[m*Pi]*Sin[n*Pi])/(m^2 - n^2) In the limit n->m we get In[7]:= Limit[%, n -> m] Out[7]= Pi/2 - Sin[2*m*Pi]/(4*m) Nice result but no Kronecker contrary to your statement. By the way, the same result appears without Assumptions. Regards, Wolfgang In[2]:= Integrate[Sin[n*x]*Sin[m*x], {x, 0, Pi}] Out[2]= (n*Cos[n*Pi]*Sin[m*Pi] - m*Cos[m*Pi]*Sin[n*Pi])/(m^2 - n^2) the same result appears using Integrate[Sin[n*x]*Sin[m*x], {x, 0, Pi}, Assumptions -> Element[{n, m}, Integers]]

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