# How does analytic continuation lets us extend functions to the complex plane?

I'm trying to understand analytic continuation and I noticed on wolfram that it

allows the natural extension of the definition trigonometric, exponential, logarithmic, power, and hyperbolic functions from the real line $\mathbb{R}$ to the entire complex plane $\mathbb{C}$

So how does it extend, say, $f(x) = \sin(x)$, $x \in \mathbb{R}$ to the complex plane? What are the steps that have to be taken to extend this function (and others) to the complex plane?

• The real line is a subset of the complex plane. If $F(z)$ is analytic on a open domain $\mathbb{D}$ that contains the real line, and is equal to $f(x)$ on the real line, then $F(z)$ is the AC of $f$. Therefore, $F(z)=\sin(z)$ is the AC of $\sin(x)$. – Mark Viola Jun 14 '16 at 16:35

More interesting is the case that Riemann worked out:

$$\Gamma(s)\zeta(s)=\int_0^\infty\frac{x^s}{e^x-1}\frac{dx}{x},$$ where $$\Gamma(s)$$ is the gamma function, $$\zeta(s)$$ is the zeta function and $$s\in\mathbb{C}.$$ To extend this formula to $$\mathbb{C},$$ Riemann considers the path integral, on the complex plane, $$\oint_C\frac{(-z)^s}{e^z-1}\frac{dz}{z},$$ where the path $$C$$ goes from $$\infty$$ to the origin $$O$$, above the positive real axis, and circulating $$O$$, counterclockwise through a circumference of radius $$\delta$$, say, returning to $$\infty$$ along the bottom of the positive real axis. The important thing, for the evaluation of above integral, is that we may split it into three integrals, namely

$$\biggl(\int_\infty^\delta+\int_{|z|=\delta}+\int_\delta^\infty\biggr)\frac{(-z)^s}{e^z-1}\frac{dz}{z},$$ recalling that $$(-z)^s=e^{s\log(-z)}$$, and $$\log(-z)=\log|z|+ i\, \text{arg}(-z).$$ So that, in the first integral, when $$-z$$ lies on the negative real axis, we take $$\text{arg}(-z)=-\pi\,i;$$ on the second one $$-z=-\delta,$$ and as $$-z$$ progress counterclockwise about $$O$$, $$\text{arg}(-z)$$ goes from $$-\pi\,i$$ to $$\pi\,i.$$ Finally, in the last integral $$\text{arg}(-z)=\pi\,i,$$ therefore the first and third integrals do not cancel. The second integral cancels as $$\delta\to 0.$$ The rest is purely technical and leads to the analytical continuation of the $$\zeta(s)$$ function of Riemann everywhere except 1 where it has a simple pole with residue 1.

There are various ways to extend a function $f:\>J\to{\mathbb R}$ defined on some open interval $J\subset{\mathbb R}$ to a holomorphic function $\tilde f:\>\Omega\to{\mathbb C}$, if such an extension is at all possible. But whatever method you use, you will always end up with the same function $\tilde f$.

(i) If $f$ is given as a rational expression $f(x)={p(x)\over q(x)}$ with $p$ and $q$ polynomials, $q(x)\ne0$ on $J$, then the extension is simply $$\tilde f(z):={p(z)\over q(z)}\ ,$$ and is defined on all of ${\mathbb C}$, minus the complex zeros of $q$.

(ii) If $f$ is given by some power series with positive radius of convergence $\rho$: $$f(x)=\sum_{k=0}^\infty c_k(x-a)^k\qquad(|x-a|<\rho)$$ then you may again put $$\tilde f(z):=\sum_{k=0}^\infty c_k(z-a)^k\qquad(|z-a|<\rho)$$ and can be sure that $\tilde f$ is holomorphic at least in the disk $D_\rho(a)$.

(iii) If $f$ is the solution of some IVP $y'=g(x,y)$, $y(x_0)=y_0$, and $g$ is "real analytic" (example: $y'=\lambda y$) then this ODE can be viewed at also in a complex sense (i.e., $x$ and $y$ are complex variables). The IVP then defines a holomorphic function $\tilde f$ in the neighborhood of $(x_0,y_0)$, which is the extension of $f$ to the complex domain.

An example: Euler's formula gives $$\sin t={e^{it}-e^{-it}\over 2i}\ .$$ We can then say right away that $$\sin z:={e^{iz}-e^{-iz}\over 2i}$$ is the complex extension of $\sin$.