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
- References:
- Re: Contour Integration
- From: dmilioto@sw.rr.com (Dominic Milioto)
- Re: Contour Integration