As pointed in Does the analytic continuation always exists? we know it doesn't always exist.
But: take the $\Gamma$ function: the first definition everyone meet is the integral one:
$$
z\mapsto\int_{0}^{+\infty}t^{z-1}e^{-t}\,dt
$$
which defines an holomorphic function on the half plane $\{\Re z>0\}$.
Moreover we immediately get the functional equation:
$$
\Gamma(z+1)=z\Gamma(z)\;,\;\;\;\forall\; \Re z>0.
$$
This equation is used to extend the function on the whole complex plane (minus the negative integers)… but: WHY CAN WE DO THIS?!
We know that there is an holomorphic function $\Gamma$ which can be expressed as the integral on that half plane. Why are we allowed to write
$$
\Gamma\left(\frac12\right)=-\frac12\Gamma\left(-\frac12\right)
$$
for example? LHS is defined, RHS, NOT!!! But where's the problem? Simply let's define $\Gamma\left(-\frac12\right)$ in such a way… but why can we do this? How can I know that this function I named $\Gamma$ which is holomorphic on the above half plane admits an extension?
Best Answer
Once you have the functional equation for $\Gamma$, we can define a new function $\tilde \Gamma$ defined on the half-plane $\operatorname{Re} z > -1$ (except $z=0$) by $$ \tilde \Gamma(z) = \frac1z \Gamma(z+1). $$ It's clear that $\tilde \Gamma$ is holomorphic on $\{ \operatorname{Re} z > -1 \} \setminus \{ 0 \}$, and coincides with $\Gamma$ on $\operatorname{Re} z > 0$ (because of the functional equation). In other words, $\tilde \Gamma$ is an analytic continuation of $\Gamma$ to $\{ \operatorname{Re} z > -1 \} \setminus \{ 0 \}$. So we may as well call $\tilde \Gamma$ by $\Gamma$.
Repeating the above construction, we can define a "new $\Gamma$-function" on successively larger sets until we get something defined and holomorphic on the whole complex plane except the non-positive integers.