# Category Archives: PolyGamma

$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$ $$\textit{proof}$$ Note that $$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$ By differentiating with respect to $a$ , $p$ times we have $$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx$$ $$\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$ Let $a =k$ $$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$ Use the relation to polygamma $$H^{(p)}_k … Continue reading Posted in Euler sum, PolyGamma, Polylogarithm | | Leave a comment ## Relation between polygamma and Hurwitz zeta function proof $\forall \,\, n\geq 1$$$\psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z)\textit{proof}$$Use the series representation of the digamma$$\psi_{0}(z) = -\gamma-\frac{1}{z}+ \sum_{n=1}^\infty\frac{z}{n(n+z)}$$This can be written as the following$$\psi_{0}(z) = -\gamma + \sum_{k=0}^\infty\frac{1}{k+1}-\frac{1}{k+z} By differentiating with respect to … Continue reading

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