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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
>>
>>
>>
>>
>>
> 
> 



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