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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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