For $\alpha\in\mathbb{C}$, I defined the "complex-weighted" Hurwitz zeta function \begin{eqnarray*}\displaystyle \zeta^{\alpha}(s,w)=\frac{1}{\Gamma(s)}\int_0^{\infty} \frac{e^{-wt}}{(1-e^{-t})^{\alpha}}t^{s-1}\,dt.\end{eqnarray*}

From this definition, we can consider a generalization of the gamma function \begin{eqnarray*}\displaystyle \Gamma^{\alpha}(w)=\exp\left(\left.\frac{\partial}{\partial s}\zeta^{\alpha}(s,w)\right|_{s=0}\right).\end{eqnarray*} In particular, $\Gamma^1(w)=\Gamma(w)/\sqrt{2\pi}$. Then I have three questions:

(1) The following representation is true? \begin{eqnarray*}\displaystyle \zeta^{\alpha}(s,w)=\sum_{n=0}^{\infty} \frac{(\alpha)_n}{n!}(n+w)^{-s},\end{eqnarray*} where $(\alpha)_n=\alpha\cdots(\alpha+n-1)$ is the Pochhammer symbol.

(2) If we could get a representation \begin{eqnarray*}\displaystyle \zeta^{\alpha}(s,w)=\frac{1}{\Gamma(s)(A(\alpha)e^{2\pi is}-1)}\int_{I(\lambda,\infty)} \frac{e^{-wt}}{(1-e^{-t})^{\alpha}}t^{s-1}\,dt,\end{eqnarray*} then what is $A(\alpha)$ ? $I(\lambda, \infty)$ is the path consisting of the infinite line from $\infty$ to $\lambda$, the circle of radius $\lambda$ around $0$ in the positive sense and the infinite line from $\lambda$ to $\infty$

(3) Is there any paper about $\zeta^{\alpha}$ or $\Gamma^{\alpha}$ ?


1 Answer 1


For $\Re(\alpha) <-1,w > 0$ then $F_{\alpha,w}(s)=\int_0^\infty t^{-\alpha} (\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt}t^{s-1}dt$ is analytic for $\Re(s) \ge 0$.

For most $\alpha,w$, $F_{\alpha,w}(0) \ne 0$ and $\zeta^\alpha(s,w)=\frac{F_{\alpha,w}(s)}{\Gamma(s)}$ has a pole at $s=0$ so it doesn't make sense to look at its derivative.

The same holds for $\Re(\alpha)\ge -1$ except we need to continue the integral analytically from $$\zeta^\alpha(s,w) - \sum_{m=0}^M c_m(\alpha,w) \frac{1}{\Gamma(s)(s+m)}= \int_0^\infty t^{-\alpha} ( (\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt} - \sum_{m=0}^M c_m(\alpha,w) t^m 1_{t < 1}) t^{s-1}dt$$ where $(\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt} = \sum_{m \ge 0}c_m(\alpha,w) t^m$ and $M > \Re(\alpha)$.

$(\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt}$ is analytic and rapidly decreasing near $[0,\infty)$ thus for $\Re(s) > 0$ $$\int_I (\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt}t^{s-1}dt$$ $$ = \int_0^\infty (\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt}t^{s-1}dt-\int_0^\infty (\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt}(e^{-2i\pi}t)^{s-1}dt$$ $$ = (1-e^{-2i \pi (s-1)})\int_0^\infty (\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt}t^{s-1}dt$$

where $I$ is the contour $\infty \to \infty$ enclosing positively $[0,\infty)$.

The contour integral $\int_I (\frac{1-e^{-t}}{t})^{-\alpha} e^{-wt}t^{s-1}dt$ is entire in $s$.

For $\Re(s) > |\alpha|$ by absolute convergence $$\int_0^\infty (1-e^{-t})^{-\alpha} e^{-wt}t^{s-1}dt= \sum_{n=0}^\infty {-\alpha \choose n} \int_0^\infty e^{-(w+n)t} t^{s-1}dt=\sum_{n=0}^\infty {-\alpha \choose n} (w+n)^{-s}\Gamma(s)$$


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