Reflection formula for Hurwitz Zeta function? In doing some calculus with Mathematica today, I found that
$$\zeta\left(3,\frac{1}{4}\right) - \zeta\left(3,\frac{3}{4}\right) = 2\pi^3$$
by numerically evaluating both sides. Here, $\zeta(x,y)$ denotes the Hurwitz Zeta function defined by
$$\zeta(x,y) = \sum_{n=0}^\infty \frac{1}{(y+n)^x},$$
called Zeta[.,.] in Mathematica. I would be very grateful if there's anyone who could provide some ideas for a proof of that or even a complete proof. I tried to find a kind of reflection theorem for this function but I didn't find anything, neither by hand nor by looking into the literature.
Thank you very much!
 A: Note
\begin{eqnarray}
&&\zeta\left(3, \frac{1}{4}\right) - \zeta\left(3, \frac{3}{4}\right)\\
&=& 2^{6} \left[ \sum_{n=0}^{\infty} \frac{1}{(4n+1)^{3}} - \sum_{n=0}^{\infty} \frac{1}{(4n+3)^{3}} \right] \\
&=& 2^{6}\sum_{n=-\infty}^{\infty} \frac{1}{(4n+1)^{3}}\\
&=&2^6(-\pi)\text{Res}\left(\frac{1}{(4z+1)^{3}}\cot(\pi z),-\frac14\right)\\
&=&2^6(-\pi)\left(-\frac{\pi^2}{32}\right)\\
&=&2\pi^3.
\end{eqnarray}
Here we used the following result from complex analysis
$$ \sum_{n=-\infty}^{\infty}f(n)=-\pi\sum_{k=1}^m\text{Res}(f(z)\cot(\pi z),a_k) $$
where $a_k$ ($k=1,2,\cdots,m$) are all poles of $f(z)$.
A: First define the $\beta(x)$ function by
\begin{align}
\beta(x) = \frac{1}{2} \, \left[ \psi\left(\frac{x+1}{2}\right) - \psi\left(\frac{x}{2}\right) \right] = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{x+k}
\end{align}
for which after two derivatives with respect to x and setting $x = 1/2$ the result becomes
\begin{align}
16 \, \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{3}} &= \frac{1}{8} \left[ \psi^{(2)}\left(\frac{x+1}{2}\right) - \psi^{(2)}\left(\frac{x}{2}\right) \right] \\
&= \frac{1}{8} \left[ 2(\pi^{3} - 28 \, \zeta(3)) - (-2)(\pi^{3} + 28 \, \zeta(3)) \right]  = \frac{\pi^{3}}{2}
\end{align}
and finally
\begin{align}
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{3}} = \frac{\pi^{3}}{32}.
\end{align}
Now,
\begin{align}
\zeta\left(3, \frac{1}{4}\right) - \zeta\left(3, \frac{3}{4}\right) &= 2^{6} \left[ \sum_{n=0}^{\infty} \frac{1}{(4n+1)^{3}} - \sum_{n=0}^{\infty} \frac{1}{(4n+3)^{3}} \right] \\
&= 2^{6} \, \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}} = 2^{6} \, \frac{\pi^{3}}{2^{5}} \\
&= 2 \, \pi^{3}. 
\end{align}
