Re: Assuming and Integrate
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- Subject: [mg128350] Re: Assuming and Integrate
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Tue, 9 Oct 2012 00:37:07 -0400 (EDT)
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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
- Assuming and Integrate