Whittaker and Watson show a derivation of the Hurwitz representation of the Hurwitz Zeta function as a trigonometrical series.
This represenation is achived by doing a countour integration. The figure below shows a bit of the definition of the contour.
The derived equation is:
\begin{eqnarray} \zeta(s,x) = \frac{2 \Gamma(1-s)}{(2 \pi)^{1-s}} \sum_{k=1}^{\infty} \frac{1}{k^{1-s}} \sin \left ( \frac{\pi s}{2} + 2 \pi k x \right ) \end{eqnarray} where I am using "$x$" instead of "$z$".
I claim that this series is the Fourier series of the Hurwitz Zeta function. I will set the task of showing this, but if anyone knows or had done this excercise I would like to see his/her development.
If I am right, it is interesting to see how analytic continuation and Fourier series are closely related.
Thanks.
Here is what I found: That is, the Fourier series of a function $f(x)$ is defined by the equations \begin{eqnarray*} f(x) = \sum_{j=-\infty}^{\infty} c_j \mathrm{e}^{\frac{2 \pi \mathrm{i} j x}{T}} \end{eqnarray*} with $c_j$ complex coefficients defined by \begin{eqnarray*} c_j &=& \frac{1}{T} \int_{0}^{T} f(t) \mathrm{e}^{\frac{-2 \pi \mathrm{i} j t}{T}} dt. \nonumber \end{eqnarray*}
We need to evaluate the coefficients $c_j$ in order to find the Fourier series representation. We need to define an interval of periodicity. We then want to force the function $\zeta(s,x)$ to be periodic on $x$. We impose $0 < x < 1$ as Hurwitz did, periodic on that interval, but we ignore what happens at the ends of the interval. We evaluate the integral in the coefficients as limits, since we do not care about the definition of $\zeta(s,x)$ at $x=0,1$. Actually if $x=1$ the Hurwitz Zeta function becomes the Riemann Zeta function. We set then $T=1$ and
\begin{eqnarray*} c_j &=& \lim_{a \to 0 } \int_{a}^{1} \zeta(s,t) \mathrm{e}^{-2 \pi \mathrm{i} j t} dt. \\ &=& \lim_{a \to 0 } \int_{a}^{1} \sum_{n=0}^{\infty} \frac{1}{(n+t)^s} \mathrm{e}^{-2 \pi \mathrm{i} j t} dt \\ &=& \lim_{a \to 0 } \sum_{n=0}^{\infty} \int_{a}^{1} \frac{1}{(n+t)^s} \mathrm{e}^{-2 \pi \mathrm{i} j t} dt \\ \end{eqnarray*}
Let us consider one term on the sum. For example choose $n=N$, and evaluate
\begin{eqnarray*} \lim_{a \to 0} \int_a^1 \frac{1}{N+t} \mathrm{e}^{-2 \pi \mathrm{i} j t } dt = \lim_{a \to 0} \int_{a+N}^{a+N+1} \frac{1}{u^s} \mathrm{e}^{-2 \pi \mathrm{i} j (u-N) } du = \int_N^{N+1} \frac{\mathrm{e}^{-2 \pi \mathrm{i} j (u-N) } }{u^s} du = \int_N^{N+1} \frac{\mathrm{e}^{- 2 \pi \mathrm{i} j u}}{u^s} du, \end{eqnarray*} where we did the change of variables $u=N+t$, $du = dt$. We claim that we can interchange the sum with the integral since the Zeta function is analytic for $x$ in the open interval $(0,1)$, and $\mathrm{Re}(s) < 1$. We then write
\begin{eqnarray*} c_j &=& \lim_{a \to 0 } \int_{a}^{1} \zeta(s,t) \mathrm{e}^{-2 \pi \mathrm{i} j t} dt = \int_{0}^{\infty} \frac{\mathrm{e}^{- 2 \pi \mathrm{i} j u}}{u^s} du \end{eqnarray*} Now we observe a similarity with the Gamma function and this suggest the change of variable $x = 2 \pi \mathrm{i} j u$, $dx = 2 \pi \mathrm{i} j du$ so
\begin{eqnarray*} c_j &=& \int_0^{\infty} \frac{\mathrm{e}^{-x}}{x^s} \frac{(2 \pi \mathrm{i} j)^s}{ 2 \pi \mathrm{i} j} dx = (2 \pi \mathrm{i} j)^{s-1} \Gamma(1-s). \end{eqnarray*}
With this
\begin{eqnarray*} \zeta(s,x) = \sum_{j=-\infty}^{\infty} (2 \pi \mathrm{i} j)^{s-1} \Gamma(1-s) \mathrm{e}^{2 \pi \mathrm{i} j x} \end{eqnarray*} since $\mathrm{Re}(s)>1$ then for $j=0$, $(2 \pi \mathrm{i} 0)^{s-1} = 0$, so we do not have the term $j=0$ in the sum. We can write the sum as
\begin{eqnarray*} \zeta(s,x) &=& (2 \pi \mathrm{i})^{s-1} \Gamma(1-s) \sum_{j=-\infty}^{\infty} j^{s-1} \mathrm{e}^{2 \pi \mathrm{i} j x} \\ &=& (2 \pi \mathrm{i})^{s-1} \Gamma(1-s) \sum_{j=1}^{\infty} j^{s-1} \mathrm{e}^{2 \pi \mathrm{i} j x} + (-j)^{s-1} \mathrm{e}^{-2 \pi \mathrm{i} j x} \\ &=& (2 \pi \mathrm{i})^{s-1} \Gamma(1-s) \sum_{j=1}^{\infty} j^{s-1} \left ( \mathrm{e}^{2 \pi \mathrm{i} j x} + (-1)^{s-1} \mathrm{e}^{-2 \pi \mathrm{i} j x} \right ) \\ &=& (2 \pi \mathrm{i})^{s-1} \Gamma(1-s) \sum_{j=1}^{\infty} j^{s-1} \left ( \mathrm{e}^{2 \pi \mathrm{i} j x} + \mathrm{e}^{-\mathrm{i} \pi (s-1)} \; \mathrm{e}^{-2 \pi \mathrm{i} j x} \right ) \\ &=& (2 \pi \mathrm{i})^{s-1} \Gamma(1-s) \mathrm{e}^{-\mathrm{i} \pi (s-1)/2} \sum_{j=1}^{\infty} j^{s-1} \left ( \mathrm{e}^{2 \pi \mathrm{i} j x + \mathrm{i} \pi(s-1)/2} + \mathrm{e}^{-2 \pi \mathrm{i} j x - \mathrm{i} \pi (s-1)/2} \right ) \\ &=& (2 \pi \mathrm{i})^{s-1} \Gamma(1-s) \mathrm{e}^{-\mathrm{i} \pi (s-1)/2} \sum_{j=1}^{\infty} 2 j^{s-1} \cos [2 \pi j x + \pi (s-1)/2 ] \\ &=& 2 (2 \pi \mathrm{i})^{s-1} \Gamma(1-s) \mathrm{e}^{-\mathrm{i} \pi (s-1)/2} \sum_{j=1}^{\infty} j^{s-1} \sin \left ( 2 \pi j x + \frac{\pi s}{2} \right ) \\ &=& \frac{2 \Gamma(1 -s)}{(2 \pi)^{1-s}} \sum_{j=1}^{\infty} \frac{1}{j^{1-s}} \sin \left ( 2 \pi j x + \frac{\pi s}{2} \right ) \end{eqnarray*} In the last step we used $\mathrm{i}^{s-1}= \mathrm{e}^{\mathrm{i} \pi(s-1)/2}$.
This is the same result obtained by Hurwitz analytic continuation by Hankel contour integration.