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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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