in terms of polylogarithms Li$_j(z)=\sum_{k=1}^\infty z^k/k^j$ for $|z| \leq 1$ and otherwise analytically continued, by making use of the following.

[**Lemma 1**]{}. We have $$\int_1^x \ln(t+1){{\ln^{j-1} t} \over t}dt=\sum_{\ell=0}^{j-1}(-1)^{\ell-1}
\ell! {{j-1} \choose \ell}\ln^{j-\ell-1} x ~\mbox{Li}_{\ell+2}(-x)+(-1)^j(j-1)!
(1-2^{-j})\zeta(j+1).  \eqno(14)$$ To prove the Lemma, we make use of the representation [@grad] (p. 568) $$\int_0^1 \ln^r x \ln(1-bx) {{dx} \over x}=(-1)^{r-1} \Gamma(r+1)\mbox{Li}_{r+2}
(b), ~~~~~~~~\mbox{Re} ~r >-1,\eqno(15)$$ where $\Gamma$ is the Gamma function, and the relation [@sri] $$\mbox{Li}_{j+1}(-1)=-(1-2^{-j})\zeta(j+1).  \eqno(16)$$ We write $$\int_1^x \ln(t+1){{\ln^{j-1} t} \over t}dt=\int_0^x \ln(t+1){{\ln^{j-1} t} 
\over t}dt -\int_0^1 \ln(t+1){{\ln^{j-1} t} \over t}dt$$ $$=\int_0^1 \ln(1+xw)(\ln x+\ln w)^{j-1} {{dw} \over w}-(-1)^j(j-1)!
\mbox{Li}_{j+1}(-1).    \eqno(17)$$ We then binomially expand the integrand in this equation and apply both Eqs. (15) and (16) and the Lemma follows.

From Lemma 1 we may now write for Eq. (12) $$\sum_{k=2}^x [\ln k-\ln(k+1)]\ln^j k=-\ln([x]+1)\ln^j x$$ $$+j\sum_{\ell=0}^{j-1}
(-1)^{\ell-1}\ell! {{j-1} \choose \ell}\ln^{j-\ell-1} x \mbox{Li}_{\ell+2}(-x)
+(-1)^jj!(1-2^{-j})\zeta(j+1)+jc_j(x).  \eqno(18)$$ From the relation between Li$_n(-x)$ and Li$_n(-1/x)$ (e.g., [@sri], p. 116) $$\mbox{Li}_n(-x) +(-1)^n\mbox{Li}_n\left(-{1 \over x}\right)=-{1 \over {n!}}
\ln^n x+2\sum_{k=1}^{[n/2]} {{\ln^{n-2k} x} \over {(n-2k)!}} \mbox{Li}_{2k}
(-1),  \eqno(19)$$ we may read off the asymptotic form $$\mbox{Li}_n(-x) \sim -{1 \over {n!}}
\ln^n x+2\sum_{k=1}^{[n/2]} {{\ln^{n-2k} x} \over {(n-2k)!}} \mbox{Li}_{2k}
(-1), ~~~~~~~~x \to \infty.  \eqno(20)$$ Then inserting this expression into the sum 