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?
 A: 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):
$$
\Im \psi\left(\frac{1}{2}+iy\right)=\frac{\pi}{2}\tanh (\pi y).
$$
