Is there a closed form expression for $\sum\limits_{k=1}^\infty \log\left(1+\frac{z^2}{k^2}\right),\ \Re(z)>0$ Question: Is there a closed form expression for $f(z)$ defined in formula $(1)$ below?
$$f(z)=\sum\limits_{k=1}^\infty \log\left(1+\frac{z^2}{k^2}\right),\quad\Re(z)>0\tag{1}$$
I believe the sum in formula $(1)$ above converges with no branches for $\Re(z)>0$. I tried simplifying this sum to a closed form representation using Mathematica but was unsuccessful.
 A: $$f(z)=\sum\limits_{k=1}^\infty \log\left(1+\frac{z^2}{k^2}\right)=\log \Bigg[\prod\limits_{k=1}^\infty \left(1+\frac{z^2}{k^2}\right)\Bigg]=\log \left(\frac{\sinh (\pi  z)}{\pi  z}\right)$$
A: We have the product representation of the Gamma function
$$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{k\ge1}e^{z/k}\left(1+\frac{z}{k}\right)^{-1},$$
which lends itself to the series
$$\log\Gamma(z)=-\gamma z-\log z+\sum_{k\ge1}\left[\frac{z}{k}-\log\left(1+\frac{z}{k}\right)\right].$$
Thus
$$\begin{align}
\log\Gamma(z)+\log\Gamma(-z)=&-\gamma z-\log z+\sum_{k\ge1}\left[\frac{z}{k}-\log\left(1+\frac{z}{k}\right)\right]\\
&+\gamma z-\log(-z)+\sum_{k\ge1}\left[-\frac{z}{k}-\log\left(1-\frac{z}{k}\right)\right]\\
=& -\log(z)-\log(-z)-\sum_{k\ge1}\left[\log\left(1+\frac{z}{k}\right)+\log\left(1-\frac{z}{k}\right)\right]\\
=&-\log(-z^2)-\sum_{k\ge1}\log\left(1-\frac{z^2}{k^2}\right).
\end{align}$$
Hence
$$\log\Gamma(iz)\Gamma(-iz)=-\log(z^2)-\sum_{k\ge1}\log\left(1+\frac{z^2}{k^2}\right),$$
so
$$f(z)=-\log(z^2)-\log\Gamma(iz)\Gamma(-iz),$$
which can likely be simplified using the Gamma reflection formula.
