= Solution
<Periodicity> reduces all the indicated vertical segments to the <compact> segment $1/2+i[-1,1]$, which misses every pole. <Continuity> therefore gives the first uniform bound. In fact an explicit identity gives a stronger estimate for every real $y$:
$$
g(m+1/2+iy)=-i\pi\tanh(\pi y),\qquad |g(m+1/2+iy)|\leq\pi.
$$
Thus we can take $D_1=\pi$.
Write $L=m+1/2$. On the vertical sides of the square the same bound holds, since both real coordinates are half-integers. On either horizontal side $|\operatorname{Im}z|=L\geq1/2$. For $\operatorname{Im}z=y>0$, put $w=e^{2\pi iz}$, so $|w|=e^{-2\pi y}$. Using
$$
\cot(\pi z)=i\frac{w+1}{w-1}
$$
gives $|g(z)|\leq\pi(1+e^{-2\pi y})/(1-e^{-2\pi y})\leq\pi\coth(\pi/2)$. <Conjugation> gives the same estimate below the real axis. Hence an explicit bound valid on every square is
$$
\boxed{D_2=\pi\coth(\pi/2).}
$$
For $L>|b|$, the <contour> has length $8L$ and every point on it has $|z|\geq L$. The <reverse triangle inequality> yields $|z^2-b^2|\geq L^2-b^2$, so the <ML inequality> gives
$$
\left|\int_{\gamma_m}\frac{g(z)}{z^2-b^2}\,dz\right|
\leq\frac{8LD_2}{L^2-b^2}\longrightarrow0.
$$
Orient the square positively. Its enclosed poles at the integers $-m,\ldots,m$ have <residues> $1/(n^2-b^2)$. Since $b$ is nonintegral, the additional poles $b$ and $-b$ are distinct from those integers and from one another. Their <residues> are $g(b)/(2b)$ and $g(-b)/(-2b)$, both equal to $g(b)/(2b)$ because $g$ is odd. Their combined contribution is therefore $\pi\cot(\pi b)/b$.
The <residue theorem> gives
$$
\frac1{2\pi i}\int_{\gamma_m}\frac{g(z)}{z^2-b^2}\,dz
=-\frac1{b^2}+2\sum_{n=1}^m\frac1{n^2-b^2}+\frac{\pi\cot(\pi b)}b.
$$
The series is <absolutely convergent>, since its terms are $O(n^{-2})$. Taking the limit, then including the $n=0$ term, yields
$$
\boxed{\sum_{n=0}^{\infty}\frac1{n^2-b^2}=-\frac1{2b^2}-\frac{\pi\cot(\pi b)}{2b},\qquad b\in\mathbb R\setminus\mathbb Z.}
$$
This is <cotangent contour summation>. Keeping the $n=0$ term separate until the last step determines the sign of the $1/(2b^2)$ contribution.
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