ヤコビの虚数変換式
f ( 1 4 , τ 4 ) = − i τ ∑ n = − ∞ ∞ i 2 n e ( 2 n ) 2 π i τ / 4 2 ∑ n = 1 ∞ e − ( n − 1 / 2 ) 2 π i / τ = − i τ ∑ n = − ∞ ∞ ( − 1 ) n e n 2 π i τ 2 ∑ n = 1 ∞ e − ( n − 1 / 2 ) 2 π i / τ = f ( 1 2 , τ ) {\displaystyle {\begin{aligned}f\left({\frac {1}{4}},{\frac {\tau }{4}}\right)&={\frac {{\sqrt {-i\tau }}\sum _{n=-\infty }^{\infty }{i^{2n}e^{(2n)^{2}{\pi }i{\tau }/4}}}{2\sum _{n=1}^{\infty }{e^{-(n-1/2)^{2}{\pi }i/\tau }}}}={\frac {{\sqrt {-i\tau }}\sum _{n=-\infty }^{\infty }{(-1)^{n}e^{n^{2}{\pi }i{\tau }}}}{2\sum _{n=1}^{\infty }{e^{-(n-1/2)^{2}{\pi }i/\tau }}}}=f\left({\frac {1}{2}},\tau \right)\\\end{aligned}}}