The function $\mathit{\psi }\left(p\right)\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{p}}\mathit{I}\mathit{n}\mathit{\Gamma }\left(p\right)$ is called the digamma function, and the polygama functions are defined by${\mathit{\psi }}^{\mathbf{n}}\left(p\right)\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{n}}}{\mathbf{d}{\mathbf{p}}^{\mathbf{n}}}\mathit{\psi }\left(p\right)$. [Warning: Some authors define $\mathit{\psi }\left(p\right)$as $\frac{\mathbf{d}}{\mathbf{d}\mathbf{p}}\mathit{I}\mathit{n}\left(p!\right)\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{p}}\mathit{I}\mathit{n}\mathit{\Gamma }\left(p+1\right)$ ).]

(a) Show that $\mathit{\psi }\left(p+1\right)\mathbf{=}\mathit{\psi }\left(p\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{p}}$. Hint: See (3.4).

(b) Use Problem 6 to obtain$\mathit{\psi }\left(p\right)\mathbf{\square }\mathit{I}\mathit{n}\left(p\right)\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}\mathbf{p}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{12}{\mathbf{p}}^{\mathbf{2}}}\mathbf{.}\mathbf{.}\mathbf{.}$.

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