Re: A FullSimplify Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg41099] Re: A FullSimplify Problem
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Fri, 2 May 2003 03:59:59 -0400 (EDT)
- References: <b8qolv$840$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Even if we consider only the case 0<a<1 I have still problems with the
integral,
f[a_]:=Integrate[h[x,a],{x,0,Pi}]
where the integrand is defined as
h[x_,a_]:=Log[1- 2*a*Cos[x] + a*a]
Instead of attempting to Simplify I went back one step and tried to
answer again the question: what is actually the value of the integral?
In my previous message I considered the function differentiated with
respect to and "proved" that f[a] == 2*Pi*Log[a]. Still: that derivation
exchanged several limiting procedures. Hence I wanted to see independent
confirmation.
(1) My good old Gradsteyn/Ryshik has an entry similar to the one in
question here, viz.
4.397 No. 6:
Integrate[Log[1 - 2a Cos[x] + a*a] Cos[n x], {x,0,Pi}]
= -Pi a^n /n for a^2 < 1
The right hand side looks fine (and can be confirmed numerically in
Mathematica) as long as n > 0 (1,2,3,...) but it does not help when n=0
because the limit n->0 does not exist.
(2) Now I consider the series expansion of the integrand with respect to a:
In[1] = Series[Log[1 - 2a Cos[x] + a*a], {a, 0, 5}] // FullSimplify
Out[1] = //OutputForm=
3 4
2 2 Cos[3 x] a Cos[4 x] a
-2 Cos[x] a - Cos[2 x] a - ------------- - ----------- -
3 2
5
2 Cos[5 x] a 6
------------- + O[a]
5
Hence the complete series expansion is
Log[1 - 2a Cos[x] + a*a] = -2 Sum[a^k Cos[k x]/k, {k, 1, Infinity}]
This series is absolutely convergent for 0<a<1. Integrating term by term
gives
In[2] = Integrate[Cos[k x], {x, 0, Pi]
Out[2] = //OutputForm=
Sin[k Pi]
---------
k
All terms vanish for integer k=1,2,3,...
Hence the integral vanishes. This contradicts previous results.
When I plot the integrand between 0 and Pi it also looks like the
integral being zero.
Performing NIntegrate (and specifying a numerically, e.g. a=0.4) on the
original integrand leads to an error message claiming a spurious
"strongly oscillating function". But by splitting the interval into
{0,Pi/2} and {Pi/2,Pi} NIntegrate behaves well and gives opposite
results for both parts. Hence the complete integral vanishes again.
Summing up, I didn't find confirmation but contradiction. To me up to
now this integral is a miracle (completely outside Mathematica). Despite
the derivation I presented in my last previous message I do now believe
that our integral is zero, mainly from considering the plot of the
completely well behaved integrand.
Hence: First it should be clarified analytically what the value of the
integral is, then we should give Mathematica a next try.
Regards,
Wolfgang
Andrzej Kozlowski wrote:
> This seems unlikely since Mathematica 4.2 returns
>
>
> FullSimplify[Integrate[Log[1-2a Cos[x]+a^2],{x,0,Pi}],{a>1}]
>
>
> 4 ¥ð Log[a]
>
> which is double the right answer. With the assumption a<-1 the answer
> is even worse (if returning a complex number when a real one is
> expected is "worse" than returning twice the right answer):
>
> FullSimplify[Integrate[Log[1 - 2*a*Cos[x] + a^2],
> {x, 0, Pi}], {a < -1}]
>
> Out[7]=
> Pi*(2*I*Pi + Log[1/a^2])
>
> Of course this does not prove that the unsimplified answer returned by
> Mathematica is wrong, in fact numerical tests suggest that it is
> probably right and the problem may be with FullSimplify. But I would
> not count on it.
>
>
>
> Andrzej Kozlowski
> Yokohama, Japan
> http://www.mimuw.edu.pl/~akoz/
> http://platon.c.u-tokyo.ac.jp/andrzej/
>
>
> On Wednesday, April 30, 2003, at 05:22 pm, Ersek, Ted R wrote:
>
>
>>At http://mathworld.wolfram.com/LeibnizIntegralRule.html
>>I learned that
>> Integrate[Log[1-2a Cos[x]+a^2],{x,0,Pi}]
>> = 2*Pi*Log[Abs[a]]
>>
>>Mathematica knows how to do this integral, but gives a much more
>>complicated
>>result. Can anyone explain how to use FullSimplify and other
>>transformations to show that the complicated result Mathematica gives
>>is
>>equivalent to the answer above?
>>
>>Thanks,
>> Ted Ersek
>>
>>
>>
>>
>>
>
>