Can somebody show me how to derive the two formulas below of which are asymptotic expansions of the gamma function? laplace and stirling series image

  • $\begingroup$ Expand Binet’s log gamma integrals into (divergent) Bernoulli-number series. Then exponentiate $\endgroup$
    – FShrike
    Aug 5 at 22:31
  • $\begingroup$ Does this also apply to the laplace series? $\endgroup$
    – Richie
    Aug 5 at 22:32
  • $\begingroup$ The “Laplace” series is obtained directly from the Stirling series by composing the exponential. For instance, $\exp u=1+u+(1/2)u^2+...$ and $\exp(1/12x+...)=1+1/12x+...$, you can mash the numbers yourself and see that they agree. A good exercise, probably, in series composition $\endgroup$
    – FShrike
    Aug 5 at 22:36
  • $\begingroup$ Could u show full working out in terms of an answer to this post as I'm a little bit confused sorry $\endgroup$
    – Richie
    Aug 5 at 22:40
  • $\begingroup$ I don’t have the time or the resources (typing that on my phone would be a nightmare). Binet’s log-gamma formulas are your friend. Choose the one in terms of exponentials, identify the integrand in terms of a (divergent) series in the Bernoulli numbers, integrate (pretending there are no limit problems), then you have the log-gamma function asymptotically expanded in terms of some exact things (the log of the $x^xe^{-x}$ and $\pi$ stuff) and this asymptotic series. To get gamma, you just take exp on both sides... this gives precisely the Stirling and Laplace series $\endgroup$
    – FShrike
    Aug 5 at 22:46


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