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Re: Assuming and Integrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg128374] Re: Assuming and Integrate
*From*: Roland Franzius <roland.franzius at uos.de>
*Date*: Thu, 11 Oct 2012 02:07:07 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*Delivered-to*: l-mathgroup@wolfram.com
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*References*: <20121006055312.C0D5B6908@smc.vnet.net> <k4r4en$noo$1@smc.vnet.net> <20121008062943.DD0416837@smc.vnet.net> <k50a8d$3sn$1@smc.vnet.net>
Am 09.10.2012 06:47, schrieb Murray Eisenberg:
> 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.
Thats one form of the correct general integral and gives both
In[31]:= Limit[
Integrate[Sin[n x] Sin[m x], {x, 0, \[Pi]},
Assumptions -> {n, m} \[Element] Integers && m^2 == n^2], n -> m]
Out[31]= \[Pi]/2 - Sin[2 m \[Pi]]/(4 m)
In[49]:= Limit[
Integrate[Sin[n x] Sin[m x], {x, 0, \[Pi]},
Assumptions -> {n, m} \[Element] Integers], n -> -m]
Out[49]= (-2 m \[Pi] + Sin[2 m \[Pi]])/(4 m)
In[50]:=
Assuming[{n, m} \[Element] Integers && n == m,
Integrate[Sin[n x] Sin[m x], {x, 0, \[Pi]}]]
Out[50]= \[Pi]/2
In[51]:=
Assuming[{n, m} \[Element] Integers && n == -m,
Integrate[Sin[n x] Sin[m x], {x, 0, \[Pi]}]]
Out[51]= -(\[Pi]/2)
Using the trigonometric product formula, the trival expression for this
obscuring formula
(n Cos[n \[Pi]] Sin[m \[Pi]] - m Cos[m \[Pi]] Sin[n \[Pi]])/(m^2 - n^2)
is of course the integral of the sum of Sincs with center at m=+-n
In[44]:=
b = Integrate[Sin[n x] Sin[m x], {x, 0, \[Pi]}] // TrigReduce // Apart
Out[44]= Sin[m \[Pi] - n \[Pi]]/(2 (m - n)) - Sin[m \[Pi] + n \[Pi]]/(
2 (m + n))
which is equal to
+-Pi/2 Sinc[Pi (n+-m) ]
but Mathematica behaves stubbornly with functions like Sinc (and Csc).
Generally speaking, "Mathematica" seems to have no clue about
orthogonality in Hilbert space bases as you can see trying any family
like BesselJ, LaguerreL HermiteH, LegendreP and JacobiP with the
corresponding domains and measures.
eg
In[57]:= Integrate[
HermiteH[5, x] HermiteH[6,
x] Exp[-x^2], {x, -\[Infinity], \[Infinity]},
Assumptions -> {n, m} \[Element] Integers && {n, m} > 0]
Out[57]= 0
In[56]:= Integrate[
HermiteH[n, x] HermiteH[m,x] Exp[-x^2], {x, -\[Infinity], \[Infinity]},
Assumptions -> {n, m} \[Element] Integers && {n, m} > 0]
During evaluation of In[56]:= Integrate::idiv: Integral of E^-x^2
HermiteH[m,x] HermiteH[n,x] does not converge on
{-\[Infinity],\[Infinity]}. >>
which is an obvously misleading message.
Out[56]= Integrate[
E^-x^2 HermiteH[m, x] HermiteH[n, x], {x, -\[Infinity], \[Infinity]},
Assumptions -> (n | m) \[Element] Integers && {n, m} > 0]
The missing chapters about definite integrals of orthogonal polynomials
eg from the Tables in Gradshtey/Rhyzik or Abramowitz we have always add
by hand, Help gives a look at the stub
Help: tutorial/OrthogonalPolynomials.
Since the direct verification of general orthogonality relations is a
difficult task, writing a general algorithm for the algebraic derivation
of orthogonality relations one has to implement Hilbert space methods
for eigenfunctions of symmetric differential operators.
Could be a Mathematica task for some in mathematical physicists, I assume.
--
Roland Franzius
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