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Re: Series problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg2119] Re: Series problem
  • From: "Brian J. Albright" <albright at physics.ucla.edu>
  • Date: Wed, 5 Jun 1996 01:38:32 -0400
  • Organization: UCLA Department of Physics and Astronomy
  • Sender: owner-wri-mathgroup at wolfram.com

Tommy Nordgren wrote:
> 
>         I want to expand the Cosine only in the expression:
> Cos[b x] Exp[-x^2]/(k^2+x^2) into a taylor series around 0.
> Computing the series in terms of x don't work, because the exponetial and
> the divisor will be expanded as well, when the series of the Cosine is
> multiplied by the other factors.
> Making the series expansion in terms of b don't work, because Mathematica
> can't integrate the resulting series expansion in terms of x.
> Are there any way to handle this except by introducing a new representation
> for function series.
> (The problem I'm currently interested in is finding a series for the function
> f[b_,k_] = Integrate[Cos[b x] Exp[-x^2]/(k^2+x^2),{x,-Infinity,Infinity}],
> which is valid for small b)
> --
> -------------------------------------------------------------------------
> Tommy Nordgren                    "Home is not where you are born,
> Royal Institute of Technology      but where your heart finds peace."
> Stockholm                         Tommy Nordgren - The dying old crone
> f85-tno at nada.kth.se

Hi.

Here's a quick way to get the nth term in a Taylor expansion 
in b for your function, provided that the contour you want
is the one along the real axis, and provided that k is chosen
in such a manner as to make the integrand regular along the
contour (like, e.g., k real and nonzero). 

g[n_] := Integrate[ 
		Normal[ Series[ Cos[b x] Exp[-x^2]/(k^2+x^2),{b,0,n}]],
		{x,-Infinity,Infinity}
	]

Be careful of the poles; for example, k=0 leads to a divergent 
integral.  
 
Hope this helps.

-Brian

-- 
Brian J. Albright                           
Department of Physics and Astronomy, UCLA    
albright at physics.ucla.edu
http://bohm.physics.ucla.edu/~albright

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