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How to obtain a series expansion of this function?

Posted 3 years ago

Hi, I have a function:

F[th_]:=(Sqrt[t]*(1 - 2*th - 5*th^2 + 2*th^3 + 2*(2 + t)*th^4 + (th*(3 + 2*th - (3 + 2*t)*th^2 - 2*th^3))/E^t))/
  (4*Sqrt[Pi]*(-1 + th)^4*(1 + th)^2) + (th*(4 - (1 + 8*t)*th - 4*(2 + t)*th^2 + (2 + 8*t + 4*t^2)*th^3 + 4*(1 + t)*th^4 - 
    th^5)*(-DawsonF[Sqrt[-(t/(-1 + th^2))]] + DawsonF[th*Sqrt[-(t/(-1 + th^2))]]/E^t))/
  (4*Sqrt[Pi]*(-1 + th)^4*(1 + th)^2*Sqrt[1 - th^2]);

where 0 < th < 1 and t>0. The argument th=1 is not in the domain of this function, but when th->1 the limiting form of the function is:

F[th_]:=(24 + 4*t - t^2 + E^t*(-24 + 20*t - 7*t^2 + 2*t^3))/(8*E^t*Sqrt[Pi]*t^(5/2));

I would like to obtain a series expansion of F[th] around th=1. Is there any way to do this? My attempt:

ser1 = Series[F[th], {th, 1, 3}, Assumptions -> {t > 0, th < 1}]
c0 = SeriesCoefficient[ser1, 0]

produces c0 identical to ser1, and hard to understand. As the limiting form for th->1 exists, and the function (extended to th=1) most likely possesses derivatives at th=1, one might expect an expansion into powers of (th-1), but the result is different. It is also not an (asymptotic?) expansion into powers of 1/(th-1). What I am doing wrong? What is the correct form of the expansion? Leslaw

POSTED BY: Leslaw Bieniasz
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