Use Stirling's formula to prove $\Gamma(z)=\int_0^{\infty}e^{-t}t^{z-1}dt$ I'm confused with Ahlfors' discussion on the proof of
$$\Gamma(z)=\int_0^{\infty}e^{-t}t^{z-1}dt.$$
At the end of the proof, he wants to proof that $|F/\Gamma|$ is a constant based on its bound, where $F(z)=\int_0^{\infty}e^{-t}t^{z-1}dt$. He proves that $|F/\Gamma|$ does not grow much more rapidly than $e^{\pi|y|/2}$. Since $|F/\Gamma|$ is of period 1, it can be expressed as a single-valued function of the variable $\xi=e^{2\pi iz}$. The following words make me confused. He says that "The function has isolated singularities at $\xi=0$ and $\xi=\infty$, and out estimate shows that $|F/\Gamma|$ grows at most like $|\xi|^{-1/4}$ for $\xi\rightarrow 0$ and $|\xi|^{1/4}$ for $\xi\rightarrow\infty$. It follows that both singularities are removable and hence $|F/\Gamma|$ must be reduce to a constant."
I understand he wants to use Liouville's theorem, but why are the singularities removable? $|\xi|^{1/4}$ is also unbounded. I cannot fill in the omitted details. Do I misunderstand something?
My question lies on page 206 of Ahlfors' Complex Analysis, third edition.
EDIT: The definition of $\Gamma(z)$ in Ahlfors' book is based on the Weierstrass' theorem on the  infinite product development of entire functions, i.e.,
$$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}.$$ Ahlfors introduces $\Gamma(z)$ by the simplest function with the negative integers for zeros, i.e.,
$$G(z)=\prod_1^{\infty}\left(1+\frac{z}{n}\right)e^{-z/n}.$$ Define $H(z)=G(z)e^{\gamma z}$ and define $\Gamma(z)=1/[zH(z)]$. Note that $\gamma$ is Euler's constant.
 A: The point is that a holomorphic function cannot grow in an arbitrary fashion near an isolated singularity. If the singularity is essential, then
$$M(r) = \sup \{ \lvert f(z)\rvert : \lvert z - z_0\rvert = r\}$$
grows faster than any power of $\frac{1}{r}$ as $r\to 0$. If it is a pole or removable, and $f \not\equiv 0$, then $M(r)$ behaves like an integer power of $r$, and even more,
$$m(r) = \inf \{ \lvert f(z)\rvert : \lvert z - z_0\rvert = r\}$$
grows (or decreases) at the same rate; there is a $k\in\mathbb{Z}$ and constants $0 < c < C < \infty$ such that
$$c\cdot r^k \leqslant m(r) \leqslant M(r) \leqslant C\cdot r^k$$
for all $r\in (0,r_0]$, for some $r_0 > 0$.
Thus any growth condition
$$\lvert f(z)\rvert \leqslant M\cdot \lvert z-z_0\rvert^\alpha$$
can be replaced with the growth condition
$$\lvert f(z)\rvert \leqslant \tilde{M}\cdot \lvert z-z_0\rvert^{\lceil\alpha\rceil}$$
using the smallest integer exponent not smaller than $\alpha$.
For if we have an estimate
$$\lvert f(z)\rvert \leqslant M\cdot \lvert z-z_0\rvert^\alpha$$
with $\alpha\in\mathbb{R}$, then for any $k\in\mathbb{Z}$ with $k > -\alpha$, the function
$$g(z) = (z-z_0)^kf(z)$$
satisfies an estimate
$$\lvert g(z)\rvert \leqslant M\lvert z-z_0\rvert^{k+\alpha} \xrightarrow{\lvert z-z_0\rvert\to 0} 0,$$
hence $g$ has a removable singularity at $z_0$, and
$$\tilde{g}(z) = \begin{cases} g(z) &, z \neq z_0 \\ \;0 &, z = z_0\end{cases}$$
is immediately seen to be a continuous extension of $g$. $\tilde{g}$ is then also holomorphic in $z_0$, hence has a Taylor series expansion
$$\tilde{g}(z) = \sum_{n = m}^\infty a_n (z-z_0)^n$$
in a neighbourhood of $z_0$, with $a_m \neq 0$ (which implies $m \geqslant 1$ since $\tilde{g}(z_0) = 0$).
Then, with $h(z) = \sum\limits_{n = 0}^\infty a_{n+m}(z-z_0)^n$, we obtain the representation
$$f(z) = \frac{h(z)}{(z-z_0)^{k-m}}$$
in a punctured neighbourhood $\{ z : 0 < \lvert z-z_0\rvert < \rho\}$ of $z_0$. Since $h(z_0) \neq 0$, and $h$ is continuous [holomorphic] in a neighbourhood of $z_0$, there is an $r_0 \geqslant 0$ and constants $0 < c < \lvert h(z_0)\rvert < C < \infty$ such that $c \leqslant \lvert h(z)\rvert \leqslant C$ for $\lvert z-z_0\rvert \leqslant r_0$. Then we have
$$\frac{c}{\lvert z-z_0\rvert^{k-m}} \leqslant \lvert f(z)\rvert \leqslant \frac{C}{\lvert z-z_0\rvert^{k-m}}$$
for $0 < \lvert z-z_0\rvert \leqslant r_0$.
If we look at the function $f\colon \mathbb{C}^\ast \to \mathbb{C}$ defined by
$$f(e^{2\pi iz}) = \frac{F(z)}{\Gamma(z)},$$
we have the growth condition
$$\lvert f(\xi)\rvert \leqslant M\cdot\lvert\xi\rvert^{-1/4}$$
for small enough $\lvert\xi\rvert$, and the above shows that that actually implies $\lvert f(\xi)\rvert \leqslant \tilde{M}$ near $0$ (take $k = 1$), so $0$ must be a removable singularity.
For the singularity at $\infty$, we can look at $g(\xi) = f(\xi^{-1})$, which satisfies the same growth condition near $0$ as $f$ does, hence has a removable singularity at $0$, or note that
$$\lim_{\xi\to\infty} \frac{f(\xi)}{\xi} = 0,$$
whence $f(\xi)/\xi$ has a zero at $\infty$, and therefore $f$ a removable singularity, or, after having removed the singularity at $0$, note that the Cauchy inequalities for the entire function $f$ imply that $f$ is constant.
A: Just by $\lim _{z\to a}(z-a)f(z)=0$  iff $a$ is a removable singularity.
