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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

Richie
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  • Expand Binet’s log gamma integrals into (divergent) Bernoulli-number series. Then exponentiate – FShrike Aug 05 '22 at 22:31
  • Does this also apply to the laplace series? – Richie Aug 05 '22 at 22:32
  • 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 – FShrike Aug 05 '22 at 22:36
  • Could u show full working out in terms of an answer to this post as I'm a little bit confused sorry – Richie Aug 05 '22 at 22:40
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    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 – FShrike Aug 05 '22 at 22:46
  • While you’re at it, Wikipedia’s page on the Gamma function lists an approximation due to Ramanujan in terms of a cubic and a sixth root (I think). I found it a good exercise to expand this polynomial-radical and we see that the first terms of the series match the first few terms of the “Laplace” series, up to some rescaling, explaining the quality of the approximation (and displaying Ramanujan’s genius, to come up with that!) – FShrike Aug 05 '22 at 22:48

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