The paper is from http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.384.1465&rep=rep1&type=pdf
I understand (1) - (17) equations and verified them. But when read about
The present approximation comes out in powers of $\beta = \alpha - 1$; if powers of $\alpha$ are preferred, a preliminary change of variables can be made $(x = e^u)$, so
$$\Gamma(\alpha) = \int_{-\infty}^{\infty}u^a\exp(-e^u)du \tag{18}$$ I am confused. If let $x= e^u$, then $u=lnx$. For $\Gamma(\alpha)=\int_0^{\infty}x^{\alpha-1}e^{-x}dx$, $ 0 < x < \infty$, then $-\infty<u<\infty$. $\Gamma(\alpha) = \int_{-\infty}^{\infty}(e^{u})^{\alpha-1}e^{-e^{u}}d(e^u) =\int_{-\infty}^{\infty}(e^{u})^{\alpha-1}e^{-e^{u}}e^udu = \int_{-\infty}^{\infty}(e^{u})^{\alpha}exp({e^{-u}})du$.
What's wrong with my derivation.
What's more,
A refinement of the argument leads to the usual asymptotic development,
$$\Gamma_1(\beta) = 1 + \frac{1}{12\beta} + ... \tag{19}$$
I am also confused with this expression. Any help will be appreciated, thanks very much.
In regards to your "What's more,", look at the lemma after equation $(12)$. That lemma is enough to obtain the first term of the asymptotic expansion for the Gamma function. The paper then says:
What the author means here is that by being more careful with the argument used to prove the lemma, the higher-order terms of the asymptotic (think of these as higher-order corrections) given in formula $(19)$ can be obtained.
The author did not prove equation $(19)$ in the paper (nor did they claim to).