Continuation of an asymptotic series originally defined for $z>0$ to $z<0$

Consider the well-known asymptotic expansion of the $\Gamma%$-function: $$\Gamma(x)\sim\left(\frac{x}{e}\right)^x \sqrt{\frac{2 \pi }{x}}\left[1+\frac{1}{12 x}+\frac{1}{288 x^2}-\frac{139}{51840 x^3}-\frac{571}{2488320 x^4}+\dots\right].\tag1$$ The series in the square brackets does not converge, but by truncating it, we can get an approximation that works pretty well for large $x$. Let's perform a change of variable $x=1/z$ and regroup factors. Now we have: $$\frac{\Gamma\!\left(\frac1z\right)(z\,e)^{1/z} }{\sqrt{2 \pi z}}\sim1+\frac{z}{12}+\frac{z^2}{288}-\frac{139 z^3}{51840}-\frac{571 z^4}{2488320}+\dots\tag2$$ Again, the series on the right does not converge for $z>0$. Now, I am stepping out of the rigorous realm into dangerous territory, but let's explore it and say that this series in some sense (that you, hopefully, will help me to figure out) represents the function on the left. If we plot both the function on the left and a truncated version of the series on the right (a polynomial), we will see that their graphs almost coincide for small positive $z$. If we extend the plot into negative $z$, then the polynomial (naturally) continues smoothly, but the function takes heavily oscillating complex values (not shown on the plot).

Question: Does the divergent series on the right of $(2)$ represent any smooth real-valued function for $z<0$ in the same sense as it represents the left side of $(2)$ for $z>0$? If yes, then what is that function? Does it have an explicit expression in terms of the $\Gamma$-function?

I understand that my question is vague and I apologize for that. I hope it makes some sense.

Solution 1:

Let us define the scaled gamma function via $$ \Gamma ^ * (z) = \frac{{\Gamma (z)}}{{\sqrt {2\pi } z^{z - 1/2} e^{ - z} }}. $$ It is known that $$ \Gamma ^ * (z) \sim \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{\gamma _n }}{{z^n }}} = 1 + \frac{1}{{12z}} + \frac{1}{{288z^2 }} - \frac{{139}}{{51840z^3 }} - \cdots $$ as $|z|\to +\infty$, $|\arg z|<\pi$, and $\gamma_n$ denote the so-called Stirling coefficients. It is also known that $$ \log \Gamma ^ * (z) \sim \sum\limits_{k = 1}^\infty {\frac{{B_{2k} }}{{2k(2k - 1)z^{2k - 1} }}} $$ as $|z|\to +\infty$, $|\arg z|<\pi$, and $B_k$ denote the Bernoulli numbers. Consequently, we have the formal identity $$ \exp \left( {\sum\limits_{k = 1}^\infty {\frac{{B_{2k} }}{{2k(2k - 1)}}x^{2k - 1} } } \right) = \sum\limits_{n = 0}^\infty {( - 1)^n \gamma _n x^n } . $$ Therefore \begin{align*} \sum\limits_{n = 0}^\infty {\gamma _n x^n } & = \sum\limits_{n = 0}^\infty {( - 1)^n \gamma _n ( - x)^n } = \exp \left( {\sum\limits_{k = 1}^\infty {\frac{{B_{2k} }}{{2k(2k - 1)}}( - x)^{2k - 1} } } \right) \\ & = \exp \left( { - \sum\limits_{k = 1}^\infty {\frac{{B_{2k} }}{{2k(2k - 1)}}x^{2k - 1} } } \right) = \left(\sum\limits_{n = 0}^\infty {( - 1)^n \gamma _n x^n } \right)^{-1}. \end{align*} In other words, $$ \frac{1}{\Gamma ^ * (z)} \sim \sum\limits_{n = 0}^\infty { \frac{{\gamma _n }}{{z^n }}} = 1 - \frac{1}{{12z}} + \frac{1}{{288z^2 }} + \frac{{139}}{{51840z^3 }} - \cdots $$ as $|z|\to +\infty$, $|\arg z|<\pi$.