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?
Answer
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.
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