Bourbaki

Example

Tag 08F2, permanent. Cite it as [Bourbaki, Tag 08F2].

Functions of a Real Variable, chapter V, § 2. ASYMPTOTIC EXPANSIONS

We have x1/1=exp⁡(log⁡x⋅exp⁡(1xlog⁡x))x^{1/1} = \exp \left( \log x \cdot \exp \left( \frac{1}{x} \log x \right) \right); when xx tends to +∞+\infty one has log⁡x≪x\log x \ll x, whence the asymptotic expansion of log⁡x⋅exp⁡(1xlog⁡x)\log x \cdot \exp \left( \frac{1}{x} \log x \right) with respect to the scale xα(log⁡x)βx^\alpha (\log x)^\beta:

log⁡x⋅exp⁡(1xlog⁡x)=log⁡x+(log⁡x)2x+12(log⁡x)3x2+o((log⁡x)3x2). \log x \cdot \exp \left( \frac{1}{x} \log x \right) = \log x + \frac{(\log x)^2}{x} + \frac{1}{2} \frac{(\log x)^3}{x^2} + o \left( \frac{(\log x)^3}{x^2} \right).

All the terms of this expansion, starting from the second, tend to 0; from this expansion and from the Taylor expansion eu=1+u+u2/2+o(u2)e^u = 1 + u + u^2/2 + o(u^2) on a neighbourhood of u=0u = 0 one deduces

x1/1=1+(log⁡x)2+12(log⁡x)4x+12(log⁡x)3x+o((log⁡x)3x). x^{1/1} = 1 + (\log x)^2 + \frac{1}{2} \frac{(\log x)^4}{x} + \frac{1}{2} \frac{(\log x)^3}{x} + o \left( \frac{(\log x)^3}{x} \right).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.