1
$\begingroup$

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 $\Gamma^{\alpha}$ ?

$\endgroup$
2
$\begingroup$

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 $$\int_0^\infty (1-e^{-t})^{-\alpha} e^{-wt}t^{s-1}dt= \sum_{n=0}^\infty {-\alpha \choose n} \int_0^\infty e^{-(w+n)t} t^{s-1}dt=\sum_{n=0}^\infty {-\alpha \choose n} (w+n)^{-s}\Gamma(s)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.