Sometimes you may find the notation $\mathit{F}\mathbf{(}\mathit{\varphi }\mathbf{,}\mathit{k}\mathbf{)}$ in (12.2)used when k> 1 . Allowing this notation, show that$\frac{\mathbf{1}}{\mathbf{3}}\mathit{F}\mathbf{(}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{,}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathit{F}\mathbf{(}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{4}}{\mathbf{5}}\mathbf{,}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{)}$. Hints: Using the Jacobi form of F in (12.2), write the integral which is equal to$\frac{\mathbf{1}}{\mathbf{3}}\mathit{F}\mathbf{(}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{,}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{)}$. Follow Example 3 to make a change of variable, write the corresponding integral, and verify that it is equal to$\frac{\mathbf{1}}{\mathbf{4}}\mathit{F}\mathbf{(}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{4}}{\mathbf{5}}\mathbf{,}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{)}$.

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