# Integral representation of generalized Euler sums

$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$

$$\textit{proof}$$

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

$$H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!}$$

Hence we have

$$H^{(p)}_k = \zeta(p) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Now divide by $k^q$ and sum with respect to $k$

$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$

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