Can someone either help derive or give a reference to the identities in Appendix B, page 27 of this, http://arxiv.org/pdf/1111.6290v2.pdf
Here is a reproduction of Appendix B from Klebanov, Pufu, Sachdev and Safdi's $2012$ preprint (v2) 'Renyi Entropies for Free Field Theories' (from the source at arxiv.org and hoping there is no problem citing it here…(RM)).
B Useful mathematical formulae
In this section we present some useful mathematical formulae. We begin with zeta function identities.
For $0 < a \leq 1$ we have the identity
$$\tag{B.1} \zeta(z, a) = \frac{2 \Gamma(1 – z)}{(2 \pi)^{1-z}} \left[\sin \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n^{1-z}}
+ \cos \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\sin 2 \pi a n}{n^{1-z}} \right] \,$$
Taking derivatives at $z=0, -1, -2$ gives
\begin{align}
\zeta'(-2, a) &= – \frac{1}{4 \pi^2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n^3}
– \frac{1}{4 \pi^3} \sum_{n=1}^\infty \frac{(2 \log (2 \pi n) + 2 \gamma – 3) \sin 2 \pi a n}{n^3} \,, \\
\tag{B.2}\zeta'(-1, a) &= \frac{1}{4 \pi} \sum_{n=1}^\infty \frac{\sin 2 \pi q n}{n^2}
– \frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{(\log (2 \pi n) + \gamma – 1) \cos 2 \pi a n}{n^2} \,, \\
\zeta'(0, a) &= \frac{1}{2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n}
+ \frac{1}{\pi} \sum_{n=1}^\infty \frac{(\log (2 \pi n) + \gamma) \sin 2 \pi a n}{n} \,.
\end{align}
Two other useful identities are the regularized sums
\begin{align}
\tag{B.3}\sum_{n \in \mathbb{Z}} \log \left( \frac{n^2}{q^2} + a^2 \right)
&= 2 \log \left[2 \sinh (\pi q |a|) \right] \,, \\
\sum_{n \in \mathbb{Z} + \frac 12} \log \left(\frac{n^2}{q^2} + a^2 \right)
&= 2 \log \left[2 \cosh (\pi q |a|) \right] \,.
\end{align}
These sums follow from the more general formula
$$\tag{B.4}\sum_{n \in \mathbb{Z}} \log \left( \frac{(n + \alpha)^2}{q^2} + a^2 \right)
= \log \left[2 \cosh (2 \pi q |a|) – 2 \cos (2 \pi \alpha) \right] \,.$$
This relation in turn follows from the Poisson summation formula
$$\tag{B.5} \frac{1}{ 2 \pi q} \sum_{n \in \mathbb{Z}} \hat f \left( \frac{n + \alpha}{q} \right) =\sum_{k \in \mathbb{Z}} e^{-i 2 \pi k \alpha} f(2 \pi q k) \,$$
applied to
$$\tag{B.6}\hat f(\omega) = \log \left( \omega^2 + a^2 \right) \,.$$
For $t \neq 0$ one can simply calculate the inverse Fourier transform of $\hat f$:
$$\tag{B.7}f(t) = \int_{-\infty}^\infty \frac{d\omega}{2 \pi} e^{-i \omega t} \log \left(\omega^2 +a^2 \right) = – \frac{e^{-|a|\;|t|}}{|t|} \,.$$
The case $t=0$ requires special care because the expression for $f(0)$ is divergent and requires regularization:
$$\tag{B.8}f(0) = \int_{-\infty}^\infty \frac{d\omega}{2 \pi} \log \left(\omega^2 +a^2 \right)
= -\frac{d}{ds} \int \frac{d\omega}{2 \pi} \frac{1}{\left(\omega^2 +a^2 \right)^s} \Biggr\rvert_{s=0}
= |a| \,.$$
Using $(B.6)-(B.8)$ one can show that $(B.5)$ reduces to $(B.4)$.
Best Answer
$(B.1)$ is the functional equation of the Hurwitz zeta function and Knopp and Robins' proof is available here. $$\tag{B.1}\zeta(z,a)=\frac{2\,\Gamma(1-z)}{(2\pi)^{1-z}}\left[\sin\frac {z\pi}2\sum_{n=1}^\infty\frac{\cos2\pi an}{n^{1-z}}+\cos\frac {z\pi}2\sum_{n=1}^\infty\frac{\sin2\pi an}{n^{1-z}}\right]$$
The author of your paper proposed a derivation of $(B.3)$ in reverse order since starting from $B.7)$. But let's try another derivation using the logarithm of the infinite products with $x:=\pi\,q\,|a|$ : $$\tag{1}\sinh(x)=x\prod_{k=1}^\infty \left(1+\frac {x^2}{\pi^2k^2}\right)$$ $$\tag{2}\cosh(x)=\prod_{k=1}^\infty \left(1+\frac {4\,x^2}{\pi^2(2k-1)^2}\right)$$ which may be found for example here or in the online references.
The derivation will not be direct since these products are convergent while $(B.3)$ is clearly divergent and needs some regularization. \begin{align} \frac 12\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)&=\frac 12\log\bigl(a^2\bigr)+\sum_{n=1}^\infty\log\left(\frac {n^2}{q^2}+a^2\right)\\ &=\log|a|+\sum_{n=1}^\infty\log\frac {n^2}{q^2}+\log\left(1+\frac{q^2a^2}{n^2}\right)\\ \tag{3}&=\log|a|+\sum_{n=1}^\infty\log\left(1+\frac{q^2a^2}{n^2}\right)+2\sum_{n=1}^\infty\log n-\log q\\ \end{align} The last sum at the right (as often in QFT) is heavily divergent so let's use zeta regularization to rewrite it in a finite form : $$f(z):=\sum_{n=1}^\infty\frac{\log n-\log q}{n^z}=-\zeta'(z)-\zeta(z)\log q,\quad \text{for}\ \Re(z)>1$$ (since $\;\displaystyle\frac d{dz}n^{-z}=\frac d{dz}e^{-z\log n}=-\frac{\log n}{n^z}$)
From this we deduce the 'zeta regularized sum' (using analytic extension of $f(z)$ down to $0$) : $$\sum_{n=1}^\infty\log n-\log q=\lim_{z\to 0^+}f(z)=-\zeta'(0)-\zeta(0)\log q=\frac{\log 2\pi}2+\frac 12\log q=\frac{\log 2\pi q}2$$ and get :
\begin{align} \frac 12\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)&=\log|a|+\sum_{n=1}^\infty\log\left(1+\frac{q^2a^2}{n^2}\right)+\log 2\pi +\log q\\ &=\log\left[2\,\pi q|a|\prod_{n=1}^\infty \left(1+\frac{q^2a^2}{n^2}\right)\right]\\ \\ &=\log[2\,\sinh(\pi\;q\,|a|)],\quad\text{using}\ (1)\ \text{for}\;\;q\,|a|=\frac x{\pi}\\ \end{align} Which is the first part of $(B.3)$ : $$\tag{B.3}\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)=2\;\log[2\,\sinh(\pi\;q\,|a|)]$$
I'll let you get the corresponding equation for $\cosh$.
Hoping this clarified things.