Calculate the amount of series We know that ‎$‎\sum_{k=1}^\infty ‎\frac{1}{k^2+1}‎ = ‎‎\frac{1}{2}(\pi ‎\coth‎(\pi) - 1)‎‎‎$‎‎. Now, how do we calculate the series‎ ‎$‎\sum_{k=1}^\infty ‎\frac{1}{(k + x)^2+1}‎$‎ for ‎‎$‎x\geq 0‎‎‎$?

One may write
$$
\begin{align}
 ‎\frac{2i}{(k + x)^2+1}&= ‎\frac{1}{k+x-i}-‎\frac{1}{k+x+i}
=\!\left(\frac{1}{k}-\frac{1}{k+x+i}\right)-\left(\frac{1}{k}-\frac{1}{k+x-i}\right)
\end{align}
$$ yielding

$$
\sum_{k=1}^\infty ‎\frac{1}{(k + x)^2+1}=\frac1{2i}\psi(x+1+i)-\frac1{2i}\psi(x+1-i),\qquad \text{Re}(x+1)>-1,\tag 1
$$

where we have used the digamma function which satisfy
$$
\psi(z+1)=-\gamma+\sum_{k=1}^\infty\left(‎\frac{1}{k}-‎\frac{1}{k+z} \right),\quad \text{Re}z>-1.\tag2
$$
For example, by putting $x=\frac12$ in $(1)$, using special values of $\psi$, one gets

$$
\sum_{k=1}^\infty ‎\frac{1}{\left(k + \frac12\right)^2+1}=\color{blue}{-\frac45+\frac \pi2 \tanh (\pi)}.\tag3
$$

