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MathGroup Archive 2005

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Re: Re: Contour Integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54011] Re: [mg53977] Re: Contour Integration
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Mon, 7 Feb 2005 03:13:01 -0500 (EST)
  • References: <200502050817.DAA22313@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Dominic Milioto wrote:

>I'm interested in evaluating a definite integral using complex
>contour.  The integrand is 1/(a nested radical) with no real zeros.
>It's posted in the Physics Forum at:
>
>http://www.physicsforums.com/showthread.php?t=61576
>
>(scroll down, someone presents it more precisely)
>
>Is this integral ammendable to complex contours?
>
>Thank You,
>Dominic Milioto  
>  
>

Not directly. To see this, first find the poles.

In[5]:= InputForm[poles = Solve[1/ee==0, x]]

Solve::verif: Potential solution {x -> -I}
     (possibly discarded by verifier) should be checked by hand. May require
     use of limits.

Solve::verif: Potential solution {x -> I}
     (possibly discarded by verifier) should be checked by hand. May require
     use of limits.

Out[5]//InputForm=
{{x -> (-I)/Sqrt[3]}, {x -> I/Sqrt[3]}, {x -> (-I)*Sqrt[3]}, {x -> 
I*Sqrt[3]}}

(+-I are in fact poles; one can check this using Limit.) Now check the 
series expansion at one of the poles.

In[7]:= Series[ee, {x, I/Sqrt[3],1}]

                                   1/8  5/8    -I        1/4
                7/8  3/8     3 (-1)    3    (------- + x)
         -3 (-1)    3                        Sqrt[3]
Out[7]= ------------------ - ------------------------------- +
             -I        1/4                  4
        4 (------- + x)
           Sqrt[3]
 
            3/8  7/8    -I        3/4
     27 (-1)    3    (------- + x)
                      Sqrt[3]               -I        5/4
 >    -------------------------------- + O[------- + x]
                    64                    Sqrt[3]

This tells you that it is not a proper Laurent expansion insofar as it 
requires fractional powers (it is a Puiseux series). Hence the residue 
method does not directly apply.


Daniel Lichtblau
Wolfram Research






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