Partial theta functionに関するRamanujanの公式3

Rogers-Fineの恒等式
$$\begin{array}{rl}\sum _{0\le n}\frac{(a;q{)}_n}{(b;q{)}_n}{t}^n& =\sum _{0\le n}\frac{(a,atq/b;q{)}_n}{(b;q{)}_n(t;q{)}_{n+1}}(bt{)}^n{q}^{n^2-n}(1-at{q}^{2n})\end{array}$$
において, $q\mapsto q^2,b=-a{q}^3,a\mapsto a^2{q}^2/t$としてから$t\to 0$として示される.

補題1において, $q\mapsto q^{\frac{3}{2}},a\mapsto a{q}^{-\frac{1}{2}}$として,
$$\begin{array}{rl}\sum _{0\le n}\frac{(-1{)}^n{a}^{2n}{q}^{\frac{1}{2}n(3n+1)}}{(-aq;q^3{)}_{n+1}}& =\sum _{0\le n}{a}^{3n}{q}^{\frac{3}{2}n(3n+1)}(1-a{q}^{3n+1})\\ & =\sum _{\begin{array}{c}0\le n\\ n\equiv 0(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}-\sum _{\begin{array}{c}0\le n\\ n\equiv 1(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\end{array}$$
一方, 右辺は
$$\begin{array}{rl}& \frac{1}{2}\sum _{0\le n}{a}^{3n}{q}^{\frac{3}{2}n(3n+1)}(1-a^2{q}^{6n+3})+\frac{1}{2}(\psi (a,q)-3aq\psi (a^3,q^9))\\ & =\frac{1}{2}\sum _{\begin{array}{c}0\le n\\ n\equiv 0(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}-\frac{1}{2}\sum _{\begin{array}{c}0\le n\\ n\equiv 2(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\\ & +\frac{1}{2}\sum _{0\le n}{a}^n{q}^{\frac{1}{2}n(n+1)}-\frac{3}{2}\sum _{\begin{array}{c}0\le n\\ n\equiv 1(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\\ & =\sum _{\begin{array}{c}0\le n\\ n\equiv 0(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}-\sum _{\begin{array}{c}0\le n\\ n\equiv 1(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\end{array}$$
となるので, 示される.

定理2の証明過程で用いた等式
$$\begin{array}{rl}\sum _{0\le n}\frac{(-1{)}^n{a}^{2n}{q}^{\frac{1}{2}n(3n+1)}}{(-aq;q^3{)}_{n+1}}& =\sum _{\begin{array}{c}0\le n\\ n\equiv 0(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}-\sum _{\begin{array}{c}0\le n\\ n\equiv 1(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\end{array}$$
において, $a\mapsto aq$として両辺に$aq$を掛けると,
$$\begin{array}{rl}\sum _{0\le n}\frac{(-1{)}^n{a}^{2n+1}{q}^{\frac{1}{2}(n+1)(3n+2)}}{(-a{q}^2;q^3{)}_{n+1}}& =\sum _{\begin{array}{c}0\le n\\ n\equiv 1(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}-\sum _{\begin{array}{c}0\le n\\ n\equiv 2(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\end{array}$$
を得る. 一方, 右辺は
$$\begin{array}{rl}& \frac{1}{2}\sum _{0\le n}{a}^{3n}{q}^{\frac{3}{2}n(3n+1)}(1-a^2{q}^{6n+3})-\frac{1}{2}(\psi (a,q)-3aq\psi (a^3,q^9))\\ & =\frac{1}{2}\sum _{\begin{array}{c}0\le n\\ n\equiv 0(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}-\frac{1}{2}\sum _{\begin{array}{c}0\le n\\ n\equiv 2(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\\ & -\frac{1}{2}\sum _{0\le n}{a}^n{q}^{\frac{1}{2}n(n+1)}+\frac{3}{2}\sum _{\begin{array}{c}0\le n\\ n\equiv 1(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\\ & =\sum _{\begin{array}{c}0\le n\\ n\equiv 1(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}-\sum _{\begin{array}{c}0\le n\\ n\equiv 2(\mathrm{mod}3)\end{array}}{a}^n{q}^{\frac{1}{2}n(n+1)}\end{array}$$
となるので, 示される.