Complex properties of the Hurwitz zeta function

Question

Is there any known expression for the complex or real part of the Hurwitz zeta function $\zeta(s,a)$?

Answer 1

Since $\Re\left( s \right) > 1 \wedge \Re\left( a \right) > 0$ in $\zeta\left( s,\, a \right)$ if $\alpha > -\frac{1}{2}$, we can use the definition $\zeta\left( s,\, a \right) = \sum_{k = 0}^{\infty}\left[ \frac{1}{\left( k + a \right)^{s}} \right]$ of the Hurwitz zeta function. To make it more general, I would define $a := x + y \cdot i \wedge \left\{ x,\, y \right\} :\in \mathbb{R} \wedge x > 0$ and $s \in \mathbb{R}$:

$$ \begin{align*} \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( k + a \right)^{s}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( k + x + y \cdot i \right)^{s}} \right]\\ \end{align*} $$

We can now put $k + x + y \cdot i$ in polar form ($\mathbb{C} \ni z = \left| z \right| \cdot e^{\arg\left( z \right) \cdot i}$):

$$ \begin{align*} \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( k + x + y \cdot i \right)^{s}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( \left| k + x + y \cdot i \right| \cdot e^{\arg\left( k + x + y \cdot i \right) \cdot i} \right)^{s}} \right]\\ \end{align*} $$

We know $\left| z \right| = \sqrt{\left( \Re\left( z \right) \right)^{2} + \left( \Im\left( z \right) \right)^{2}}$ and $k + x = \Re \wedge y = \Im$:

$$ \begin{align*} \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( \left| k + x + y \cdot i \right| \cdot e^{\arg\left( k + x + y \cdot i \right) \cdot i} \right)^{s}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( \sqrt{\left( k + x \right)^{2} + y^{2}} \cdot e^{\arg\left( k + x + y \cdot i \right) \cdot i} \right)^{s}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( \sqrt{\left( k + x \right)^{2} + y^{2}} \right)^{s} \cdot \left( e^{\arg\left( k + x + y \cdot i \right) \cdot i} \right)^{s}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{1}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}} \cdot e^{\arg\left( k + x + y \cdot i \right) \cdot s \cdot i}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{e^{-\arg\left( k + x + y \cdot i \right) \cdot s \cdot i}}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right]\\ \end{align*} $$

We knwo from Euler's Formula: $e^{x \cdot i} = \cos\left( x \right) + \sin\left( x \right) \implies e^{-x \cdot i} = \cos\left( x \right) - \sin\left( x \right)$

$$ \begin{align*} \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{e^{-\arg\left( k + x + y \cdot i \right) \cdot s \cdot i}}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\cos\left( \arg\left( k + x + y \cdot i \right) \cdot s \right) - \sin\left( \arg\left( k + x + y \cdot i \right) \cdot s \right) \cdot i}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\cos\left( \arg\left( k + x + y \cdot i \right) \cdot s \right)}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} - \frac{\sin\left( \arg\left( k + x + y \cdot i \right) \cdot s \right) \cdot i}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\cos\left( \arg\left( k + x + y \cdot i \right) \cdot s \right)}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right] - \sum\limits_{k = 0}^{\infty}\left[ \frac{\sin\left( \arg\left( k + x + y \cdot i \right) \cdot s \right) \cdot i}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right]\\ \zeta\left( s,\, a \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\cos\left( \arg\left( k + x + y \cdot i \right) \cdot s \right)}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right] - \sum\limits_{k = 0}^{\infty}\left[ \frac{\sin\left( \arg\left( k + x + y \cdot i \right) \cdot s \right)}{\left( \left( k + x \right)^{2} + y^{2} \right)^{\frac{s}{2}}} \right] \cdot i\\ \end{align*} $$

Both parts before the $i$ are real aka: $$\fbox{$ \begin{align*} \Re\left( \zeta\left( s,\, a \right) \right) &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\cos\left( \arg\left( k + \Re\left( a \right) + \Im\left( a \right) \cdot i \right) \cdot s \right)}{\left( \left( k + \Re\left( a \right) \right)^{2} + \left( \Im\left( a \right) \right)^{2} \right)^{\frac{s}{2}}} \right]\\ \Im\left( \zeta\left( s,\, a \right) \right) &= -\sum\limits_{k = 0}^{\infty}\left[ \frac{\sin\left( \arg\left( k + \Re\left( a \right) + \Im\left( a \right) \cdot i \right) \cdot s \right)}{\left( \left( k + \Re\left( a \right) \right)^{2} + \left( \Im\left( a \right) \right)^{2} \right)^{\frac{s}{2}}} \right]\\ \end{align*} $} \tag{1}$$

The other complex component (the complex argument) is then simply composed of the two: $$\fbox{$ \begin{align*} \arg\left( \zeta\left( s,\, a \right) \right) &= \operatorname{arctan2}\left( -\sum\limits_{k = 0}^{\infty}\left[ \frac{\sin\left( \arg\left( k + \Re\left( a \right) + \Im\left( a \right) \cdot i \right) \cdot s \right)}{\left( \left( k + \Re\left( a \right) \right)^{2} + \left( \Im\left( a \right) \right)^{2} \right)^{\frac{s}{2}}} \right],\, \sum\limits_{k = 0}^{\infty}\left[ \frac{\cos\left( \arg\left( k + \Re\left( a \right) + \Im\left( a \right) \cdot i \right) \cdot s \right)}{\left( \left( k + \Re\left( a \right) \right)^{2} + \left( \Im\left( a \right) \right)^{2} \right)^{\frac{s}{2}}} \right] \right)\\ \end{align*} $} \tag{2}$$