       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:
>
>
>(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:= 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//InputForm=
{{x -> (-I)/Sqrt}, {x -> I/Sqrt}, {x -> (-I)*Sqrt}, {x ->
I*Sqrt}}

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

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

1/8  5/8    -I        1/4
7/8  3/8     3 (-1)    3    (------- + x)
-3 (-1)    3                        Sqrt
Out= ------------------ - ------------------------------- +
-I        1/4                  4
4 (------- + x)
Sqrt

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

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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