About the series $\sum_{n\geq 0}\frac{1}{(2n+1)^2+k}$ and the digamma function Let we provide a closed form for 
$$ S_k = \sum_{n\geq 0}\frac{1}{(2n+1)^2+k} $$
for $k>0$ in terms of elementary functions. It is quite easy to check that $S_k$ can be computed in terms of the digamma function $\psi(x)$, but it is also true that:
$$ \int_{0}^{+\infty}\frac{\sin(mx)}{m}e^{-\sqrt{k}\,x}\,dx =\frac{1}{m^2+k}\tag{1} $$
and that, almost everywhere:
$$ \sum_{n\geq 0}\frac{\sin((2n+1) x)}{2n+1} = \frac{\pi}{4}(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor}, \tag{2}$$
hence:
$$\begin{eqnarray*} S_k &=& \frac{\pi}{4}\int_{0}^{+\infty}(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor} e^{-\sqrt{k}\,x}\,dx = \frac{\pi}{4}\sum_{n\geq 0}(-1)^n \int_{n\pi}^{(n+1)\pi}e^{-\sqrt{k}\,x}\,dx \\&=&\frac{\pi}{4\sqrt{k}}\sum_{n\geq 0}(-1)^n\left(e^{-n\pi\sqrt{k}}-e^{-(n+1)\pi\sqrt{k}}\right)\\&=&\frac{\pi\left(1-e^{-\pi\sqrt{k}}\right)  }{4\sqrt{k}}\sum_{n\geq 0}(-1)^n e^{-n\pi\sqrt{k}}=\color{red}{\frac{\pi}{4\sqrt{k}}\cdot\frac{e^{\pi\sqrt{k}}-1}{e^{\pi\sqrt{k}}+1}}.\tag{3} \end{eqnarray*}$$
Does this identity provide something interesting about special values of the digamma function?

Your result just confirms already known special values of the digamma function.
One may recall that the digamma function admits the following series representation 
$$\begin{equation} 
\psi(x+1) = -\gamma - \sum_{k=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}  
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant. By partial fraction decomposition we have
$$
\begin{align}
\frac{1}{(2n+1)^2+k} &= \frac{i}{2\sqrt{k}}\left(\frac{1}{(2n+1)+i\sqrt{k}}-\frac{1}{(2n+1)-i\sqrt{k}}\right)\\\\
&=\frac{i}{4\sqrt{k}}\left(\frac{1}{n+\frac{1+i\sqrt{k}}{2}}-\frac{1}{n+\frac{1-i\sqrt{k}}{2}}\right)\\\\
&=\frac{i}{4\sqrt{k}}\left[\left(\frac{1}{n+\frac{1+i\sqrt{k}}{2}}-\frac1n\right)-\left(\frac{1}{n+\frac{1-i\sqrt{k}}{2}}-\frac1n\right)\right]
\end{align}
$$ then summing from $n=1$ to $+\infty$, using $(1)$, we get
$$
\sum_{n=1}^{\infty}\frac{1}{(2n+1)^2+k} =\frac{i}{4\sqrt{k}}\left(\psi\left(\frac{1-i\sqrt{k}}{2}+1\right)-\psi\left(\frac{1+i\sqrt{k}}{2}+1\right)\right)
$$ using 
$$
\psi(x+1)=\psi(x)+\frac1x
$$ it gives the case $n=0$ to obtain
$$
\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2+k} =\frac{i}{4\sqrt{k}}\left(\psi\left(\frac{1}{2}-\frac{i\sqrt{k}}{2}\right)-\psi\left(\frac{1}{2}+\frac{i\sqrt{k}}{2}\right)\right)=\color{red}{\frac{\pi}{4\sqrt{k}}\cdot\frac{e^{\pi\sqrt{k}}-1}{e^{\pi\sqrt{k}}+1}}
$$ where we have used the result (6.3.12):
$$
\Im \psi\left(\frac{1}{2}+iy\right)=\frac{\pi}{2}\tanh (\pi y).
$$
