Analytic extension of $\sum_{k=1}^n\frac1k$ complex domain

Question

The analytic extension:

$$\sum_{k=1}^n\frac1k=\int_0^1\frac{x^n-1}{x-1}dx$$

I was wondering for what values of $n$ does this extend to, mainly complex values of $n$.

I know it is defined for $n=0$,

$$\int_0^1\frac{x^0-1}{x-1}dx=\int_0^10dx=0$$

I also know it is not defined for $n<0$

$$\int_0^1\frac{x^{-n}-1}{x-1}dx=\int_0^1\frac{-1}{x^n}dx\to\infty$$

But what about $n\in\mathbb{C}$?

Answer 1

A natural way to extend $$ \int_0^1\frac{x^n-1}{x-1}dx=\sum_{k=1}^n\frac1k, \quad n=1,2,\ldots, $$ is to consider $$ \int_0^1\frac{x^z-1}{x-1}dx=\psi(z+1)+\gamma, \quad z>-1, $$ where $\gamma$ is the Euler-Mascheroni constant and where $z \mapsto \psi(z+1)$ is the digamma function, analytic over $(-1,\infty)$, deeply linked to the Euler gamma function $$ \psi(z+1)=\frac{\Gamma'(z+1)}{\Gamma(z+1)}, \quad z>-1, $$ and which has been extensively studied since Euler's works ($1728$).

A meromorphic extension does exist, $$ \psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{z+n} \right), \quad z \in \mathbb{C}\setminus\mathbb{\left\{-1,-2,-3 \ldots \right\}}. $$