Principal logarithm of the Gamma function: an elementary approach? I have been trying to learn about the Gamma function and it's properties from various sources. As is common practice, I am defining the Gamma function by the formula
$$\Gamma(s):= \int_0^\infty e^{-t} t^{s-1} \, dt$$
for $\Re(s)>0$, and its meromorphic continuation to the whole plane by the Weierstrass product formula
$$\frac1{s\Gamma(s)} = e^{\gamma s} \prod_{n=1}^\infty (1+s/n) e^{-s/n},$$
with $\gamma$ denoting the Euler-Mascheroni constant.
In most places, I have noticed that people directly start talking about the principal branch of $\log \Gamma(s)$ (giving its series and discussing Stirling's formula) for $s \in \mathbb C \setminus \mathbb R_{\le 0}$. However, this is defined and holomorphic only when $\Gamma(s) \not\in \mathbb R_{\le 0}$ for any $s \in \mathbb C \setminus \mathbb R_{\le 0}$. It is not clear to me why this latter assertion holds except for in the obvious cases where $s>0$, and I would very much appreciate any help in that regard. I have tried seeing it as a consequence of the Weierstrass product and have also tried combining the aforementioned integral representation of $\Gamma(s)$ with the functional equation $\Gamma(s+1)=s\Gamma(s)$ but to no avail. I feel like I am missing something really straightforward or am having trouble putting the pieces together in the right manner.
 A: You don't need to know that $\Gamma(s)\not\in\mathbb{R}_{\leq 0}$ for $s\not\in\mathbb{R}_{\leq 0}$ to define the logarithm (indeed, I don't think that is even true).  Instead, you can just "formally" take a logarithm of the product formula, turning the product into a sum.  That is, take $$\Gamma(s)=\frac{e^{-\gamma s}}{s}\prod_{n=1}^\infty(1+s/n)^{-1}e^{s/n}$$ and take the logarithm factor-by-factor to define $$\log\Gamma(s)=-\gamma s -\log(s)+\sum_{n=1}^\infty(s/n-\log(1+s/n))$$ where on the right side you use the principal branch of the logarithm on $\mathbb{C}\setminus\mathbb{R}_{\leq 0}$ (which makes sense because if $s\in \mathbb{C}\setminus\mathbb{R}_{\leq 0}$ then $1+s/n\in \mathbb{C}\setminus\mathbb{R}_{\leq 0}$ as well).  The sum converges locally uniformly so this function is holomorphic, and exponentiating it gives you back $\Gamma(s)$.  Also, it is easy to see that if $s>0$ then this $\log \Gamma(s)$ is the usual real-valued logarithm, so this is the unique analytic continuation of the real-valued $\log \Gamma(s)$ function for $s>0$ to all of $\mathbb{C}\setminus\mathbb{R}_{\leq 0}$.
Alternatively, there is a general theorem that if $f$ is a nowhere vanishing holomorphic function on a simply connected domain, then $f$ has a well-defined holomorphic logarithm (defined by integrating the logarithmic derivative $f'/f$; see this answer for more details).  So in order to define $\log\Gamma$ on $\mathbb{C}\setminus\mathbb{R}_{\leq 0}$, you only need to know that it is nonvanishing on this domain, since $\mathbb{C}\setminus\mathbb{R}_{\leq 0}$ is simply connected.
