# Can somebody show me how u can derive the stirling series (with coefficients) and the laplace series of the gamma function

Can somebody show me how to derive the two formulas below of which are asymptotic expansions of the gamma function? • Expand Binet’s log gamma integrals into (divergent) Bernoulli-number series. Then exponentiate Aug 5 at 22:31
• Does this also apply to the laplace series? Aug 5 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 Aug 5 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 Aug 5 at 22:40
• 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 Aug 5 at 22:46