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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?

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  • $\begingroup$ You are just using the same name for two differenti objects. $\endgroup$ – N74 Apr 13 '16 at 20:11
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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.

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    $\begingroup$ I don't agree. you are totally obfuscating the process of analytic continuation this way : what about the analytic continuation on a path of intersecting disks ? how do you know that the analytic continuation doesn't depend on the path (the fact that $\Gamma(s)$ is meromorphic ?) $\endgroup$ – reuns Apr 13 '16 at 21:47
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The following are citations from the classic Applied and Computational Complex Analysis, Vol. I by P. Henrici. The chapter 3: Analytic Continuation provides a thorough treatise of the theme. Here we look at two aspects, which should help to clarify the situation.

At first we take a look when two functions $f(z)$ and $g(z)$ are analytic continuations from each other.

Theorem 3.2d: (Fundamental lemma on analytic continuation)

Let $Q$ be a set with point of accumulation $q$ and let $R$ and $S$ be two regions such that their intersection contains $Q$ and $q$ is connected. If $f$ is analytic on $R$, $g$ is analytic on $S$, and $f(z)=g(z)$ for $z\in Q$, then $f(z)=g(z)$ throughout $R\cap S$ and $f$ and $g$ are analytic continuations of each other.

We observe, we need at least a set $Q$ with an accumulation point where analytic functions $f$ and $g$ have to coincide. This set is part of the intersection of two regions $R$ and $S$ where $f$ and $g$ are defined. Finally we conclude that throughout $R\cap S$ the functions coincide.

The second aspect sheds some light at functional relationships in connection with analytic continuation. We can read in

Section 3.2.5: Analytic Continuation by Exploiting Functional Relationships

Occasionally an analytic continuation of a function $f$ can be obtained by making use of a special functional relationship satisfied by $f$. Naturally this method is restricted to those functions for which such relationships are known.

He continues with example 15 which seems that P. Henrici had precisely a user with OPs question in mind.

Example 15:

Let the function $g$ possess the following properties:

  • (a) $g$ is analytic in the right half-plane: $R:\Re (z)>0$

  • (b) For all $z\in R, zg(z)=g(z+1)$

We assert that $g$ can be continued analytically into the whole complex plane with the exception of the points $z=0,-1,-2,\ldots$.

We first continue $g$ into $S:\Re (z)>-1,z\neq 0$. For $z\in S$,let $f$ be defined by \begin{align*} f(z):=\frac{1}{z}g(z+1) \end{align*} For $z\in S, \Re(z+1)>0$. Hence by virtue of (a) $f$ is analytic on $S$. In view of (b) $f$ agrees with $g$ on the set of $R$. Since $S$ is a region, $f$ represents the analytic continuation of $g$ from $R$ to $S$. We note that $f$ satisfies the functional relation $f(z+1)=zf(z)$ on the whole set $S$.

Denoting the extended function again by $g$, we may use the same method to continue $g$ analytically into the set $\Re(z)>-2,z\neq 0,-1$, and thus step by step into the region $z\neq 0,-1,-2,\ldots$.

Of course this example addresses the Gamma Function $\Gamma(z)$ which is treated in detail in chapter 8, vol. 2. He then continues with further methods of analytic continuation, such as the principle of continuous continuation and the symmetry principle.

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