三角関数の部分分数展開
従って、 f ( z ) {\displaystyle f(z)} は | I z | < ∞ {\displaystyle |\Im {z}|<\infty } において有界である。 z = x + i y {\displaystyle z=x+iy} と書き
lim y → ∞ | π cot  π z | = lim y → ∞ | cos  π x cosh  π y + i sin  π x sinh  π y sin  π x cosh  π y + i cos  π x sinh  π y | ≤ lim y → ∞ | cos  π x cosh  π y + i sin  π x cosh  π y sin  π x sinh  π y + i cos  π x sinh  π y | = lim y → ∞ | cos  π x + i sin  π x sin  π x + i cos  π x | = 1 {\displaystyle {\begin{aligned}\lim _{y\to \infty }\left|\pi \cot {{\pi }z}\right|&=\lim _{y\to \infty }\left|{\frac {\cos {{\pi }x}\cosh {{\pi }y}+i\sin {{\pi }x}\sinh {{\pi }y}}{\sin {{\pi }x}\cosh {{\pi }y}+i\cos {{\pi }x}\sinh {{\pi }y}}}\right|\\&\leq \lim _{y\to \infty }\left|{\frac {\cos {{\pi }x}\cosh {{\pi }y}+i\sin {{\pi }x}\cosh {{\pi }y}}{\sin {{\pi }x}\sinh {{\pi }y}+i\cos {{\pi }x}\sinh {{\pi }y}}}\right|\\&=\lim _{y\to \infty }\left|{\frac {\cos {\pi }x+i\sin {\pi }x}{\sin {\pi }x+i\cos {\pi }x}}\right|\\&=1\\\end{aligned}}}
| ∑ n = 1 ∞ 2 z z 2 − n 2 | ≤ ∑ n = 1 ∞ | 2 ( x + i y ) x 2 − y 2 + 2 i x y − n 2 | ≤ ∑ n = 1 ∞ | 1 + 2 | y | n 2 + | y | 2 − 1 4 | ≤ ∫ n = 0 ∞ | 1 + 2 | y | n 2 + | y | 2 − 1 4 | d n {\displaystyle {\begin{aligned}\left|\sum _{n=1}^{\infty }{\frac {2z}{z^{2}-n^{2}}}\right|&\leq \sum _{n=1}^{\infty }\left|{\frac {2(x+iy)}{x^{2}-y^{2}+2ixy-n^{2}}}\right|\\&\leq \sum _{n=1}^{\infty }\left|{\frac {1+2|y|}{n^{2}+|y|^{2}-{\frac {1}{4}}}}\right|\\&\leq \int _{n=0}^{\infty }\left|{\frac {1+2|y|}{n^{2}+|y|^{2}-{\frac {1}{4}}}}\right|dn\\\end{aligned}}}
tan  θ = n | y | 2 − 1 4 {\displaystyle \tan \theta ={\frac {n}{\sqrt {|y|^{2}-{\frac {1}{4}}}}}} の置換により
| ∑ n = 1 ∞ 2 z z 2 − n 2 | ≤ 1 + 2 | y | | y | 2 − 1 4 ⋅ π 2 {\displaystyle \left|\sum _{n=1}^{\infty }{\frac {2z}{z^{2}-n^{2}}}\right|\leq {\frac {1+2|y|}{\sqrt {|y|^{2}-{\frac {1}{4}}}}}\cdot {\frac {\pi }{2}}}
π cos  π z = π sin  π ( z + 1 2 ) = lim N → ∞ ∑ n = − N N ( − 1 ) n z + 1 2 + n = − ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) z 2 − ( n + 1 2 ) 2 {\displaystyle {\begin{aligned}{\frac {\pi }{\cos {{\pi }z}}}&={\frac {\pi }{\sin {{\pi }\left(z+{\frac {1}{2}}\right)}}}\\&=\lim _{N\to \infty }\sum _{n=-N}^{N}{\frac {(-1)^{n}}{z+{\frac {1}{2}}+n}}=-\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n+1)}{z^{2}-\left(n+{\frac {1}{2}}\right)^{2}}}\\\end{aligned}}}