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Re: Assuming and Integrate

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  • Subject: [mg128350] Re: Assuming and Integrate
  • From: "Dr. Wolfgang Hintze" <weh at>
  • Date: Tue, 9 Oct 2012 00:37:07 -0400 (EDT)
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On 8 Okt., 08:52, Roland Franzius <roland.franz... at> 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 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.


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]]

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