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6

It came as a little surprise to me that in spite of the essential singularity of Exp[1/T] the expansion of the integral in powers of T about T=0 exists. In fact it can be easily written down with the asymptotic expansion of the function PolyLog[s,z] for |z|-> oo (https://en.wikipedia.org/wiki/Polylogarithm). In our case we have z = - Exp[1/T] and s = ...

3

You can fit it, if you like. It can be done as follows. First this is the integral: int = Simplify[ Integrate[ p^4/(1 + Exp[p^2/(2 m T) - \[Mu]/T]), {p, 0, \[Infinity]}], {m > 0, T > 0}] (* -3 Sqrt[\[Pi]/2] (m T)^(5/2) PolyLog[5/2, -E^((\[Mu]/T))] *) Let us change the variable: T -> t*mu: expr1 = int /. T -> t*\[Mu] (* -3 ...

2

I stumbled across this question just now, but it seems to me that there's a much easier way to do this problem via user-defined SeriesData objects. If we want to define $$f(x) = c_0 + c_1 x + c_2 x^2 + \dots$$ it's not too hard to do this in Mathematica: n = 4; f[x_] = SeriesData[x, 0, Array[c, n, 0], 0, n, 1] (* c[0] + c[1] x + c[2] x^2 + c[3] x^3 + ...

5

Using Seriesas suggested by @Hintze: Series[n!, {n, Infinity, 4}]/(Sqrt[2 Pi n] n^(n Exp[n])) // FullSimplify (*Exp[(-1 - Log[1/n]) n + SeriesData[n, DirectedInfinity[1], {}, -1, 1, 1]^5] * (SeriesData[n, Infinity, {1, 0, 1/12, 0, 1/288, 0, -139/51840, 0, -571/2488320}, 0, 10, 2]/n^(E^n*n))*) Also, Eric W. Weisstein provides some documentation on ...

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