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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 $y$: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 $|\mathrm{Im}z|=L\ge 1/2$. For $\mathrm{Im}z=y>0$, put $w=e^{2\pi iz}$, so $|w|=e^{-2\pi y}$. Usinggives $|g(z)|\le \pi (1+e^{-2\pi y})/(1-e^{-2\pi y})\le \pi \coth (\pi /2)$. Conjugation gives the same estimate below the real axis. Hence an explicit bound valid on every square isFor $L>|b|$, the contour has length $8L$ and every point on it has $|z|\ge L$. The reverse triangle inequality yields $|z^2-b^2|\ge L^2-b^2$, so the ML inequality givesOrient the square positively. Its enclosed poles at the integers $-m,\dots ,m$ 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 givesThe series is absolutely convergent, since its terms are $O(n^{-2})$. Taking the limit, then including the $n=0$ term, yieldsThis is cotangent contour summation. Keeping the $n=0$ term separate until the last step determines the sign of the $1/(2b^2)$ contribution.

For a pole of order $N$ at $z_0$, the residue is ${\mathrm{Res}}_{z_0}F=\frac{1}{(N-1)!}{\lim }_{z\to z_0}\frac{d^{N-1}}{d{z}^{N-1}}[(z-z_0{)}^{N}F(z)]$. For the rational integral, close the real-axis contour with an upper semicircle. The poles are double at $i$ and simple at $2i$, withThe arc integral vanishes as $O(R^{-5})$. By the residue theorem the whole real-axis integral is $\pi /9$, and evenness givesFor the second calculation, reciprocal expansion of the sine gives $1/\sin z=z^{-1}+z/6+O(z^3)$. Multiplying by $\cos z=1-z^2/2+O(z^4)$ yields $\cot z=z^{-1}-z/3+O(z^3)$. ThereforeAt every nonzero integer $n$, $\pi \cot \pi z$ has a simple pole with residue one, giving residue $1/n^2$ after division by $z^2$.

The cotangent contour summation uses the positively oriented square of half-side $R=N+1/2$. Its length is $8R$ and $|z|\ge R$ on its boundary. The supplied cotangent bound implies $|\oint Fdz|\le 8\pi C/R\to 0$. Its enclosed residues sum to $-{\pi }^2/3+2{\sum }_{n=1}^N{n}^{-2}$. The residue theorem therefore proves the Basel problem value

For fixed $z\notin \mathbb{Z}$, integratecounterclockwise around the specified square of half-side $R_N=N+1/2$, choosing $N>|z|$. The integer poles inside it are $-N,\dots ,N$, and their residues are $1/(n^2-z^2)$ because $\pi \cot (\pi w)$ has residue one at each integer.

The two other possible poles are $w=z$ and $w=-z$. Their residues are each $\pi \cot (\pi z)/(2z)$, hence sum to $\pi \cot (\pi z)/z$. This formula still applies when the numerator vanishes and a possible pole is removable. The residue theorem givesOn the square, $|w|\ge R_N$, its perimeter is $8R_N$, and the supplied cotangent bound givesThe series converges absolutely by part (b), so its symmetric partial sums have the same limit as the full sum. ThereforeAt half-integers the cotangent form makes the value zero explicit; the reciprocal-tangent notation is understood by its continuous extension there. The cotangent contour summation rests on the contour estimate, not only on formally matching poles.