3
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FullSimplify[Series[2 DawsonF[x], {x, ∞, 8}]]
(* -I Exp[-x^2] √π + (1/x + 1/(2 x^3) + 3/(4 x^5) + 15/(8 x^7) + O[1/x]^9) *)

What is the reason the term -I Exp[-x^2] √π is present here? The function being expanded is real-valued, and anyways the term decays faster than any of the polynomial terms given here, so why include it in the series, when I specified that I need terms up to the order 8?

The issue is reproducible both in version 8 and 10.2.

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  • $\begingroup$ FunctionExpand[DawsonF[x]] probably gives a bit of a clue $\endgroup$ – chuy Apr 22 '16 at 18:23
  • $\begingroup$ To add to chuy's comment, note that the known asymptotic expansions for the imaginary error function (see e.g. the Wolfram Functions site) necessarily involve a complex term. $\endgroup$ – J. M. will be back soon Apr 22 '16 at 22:52
  • 2
    $\begingroup$ I still do not get it. Why there should be an imaginary term, if I'm interested in the asymptotic expansion when $x$ approaches the positive infinity along the real axis? $\endgroup$ – Vladimir Reshetnikov Apr 23 '16 at 3:08

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