(k+a)} \over {k+a}}
+\sum_{k=0}^x [\ln (k+a)-\ln(k+a+1)]\ln^j (k+a)\right \}.  \eqno(25)$$ We next apply summation by parts, writing $$\sum_{k=0}^x [\ln (k+a)-\ln(k+a+1)]\ln^j (k+a)=[\ln a-\ln(a+1)]\ln^j a
+[\ln(a+1)-\ln([x]+a+1)]\ln^j(x+a)$$ $$-j\int_0^x[\ln(a+1)-\ln([t]+a+1)]{{\ln^{j-1} (t+a)} \over {t+a}}dt.  \eqno(26)$$ The last term here can be written $$-j\int_0^x[\ln(a+1)-\ln([t]+a+1)]{{\ln^{j-1} (t+a)} \over {t+a}}dt
=\ln(a+1)[\ln^j a-\ln^j(x+a)]$$ $$+j\left(\int_0^x\ln(t+a+1){{\ln^{j-1} (t+a)} \over {t+a}}dt+d_j^{(x)}(a)\right), \eqno(27)$$ where we have put $$d_j^{(x)}(a) \equiv \int_0^x {{\ln^{j-1} (t+a)} \over {(t+a)}} \ln\left({{[t]+a+1}\over {t+a+1}}\right )dt, ~~~~~~~~~~j \geq 1.\eqno(28)$$

The integral of Eq. (27) is obtained from Lemma 1 as $$\int_0^x \ln(t+a+1){{\ln^{j-1} (t+a)} \over {(t+a)}}dt
=\left(\int_1^{x+a}-\int_1^a\right)\ln(u+1) {{\ln^{j-1} u} \over u}du$$ $$=\sum_{\ell=0}^{j-1}
(-1)^{\ell-1} \ell! {{j-1} \choose \ell}\left[\ln^{j-\ell-1} (x+a) ~\mbox{Li}_{\ell+2}
(-x-a)$$ -\^[j--1]{} a  \_[+2]{}(-a)\]. (29)$$We combine Eqs. (26), (27) and (29), giving$$\_[k=0]{}\^x \[(k+a)-(k+a+1)\]\^j (k+a)=\[a-(a+1)\]\^j a +\[(a+1)-(\[x\]+a+1)\]\^j(x+a)$$$$+(a+1)\[\^j a-\^j(x+a)\]+jd\_j\^[(x)]{}(a)$$$$+j\_[=0]{}\^[j-1]{}(-1)\^[-1]{} ! [[j-1]{} ]{}. (30)$$We then make use of the asymptotic relation (20), obtaining$$\_[m=2]{}\^(-1)\^j \^[(j)]{}(m,a) =\_[x ]{}{\_[k=0]{}\^x [[\^j (k+a)]{} ]{}+\^[j+1]{} a -\^[j+1]{} (x+a)+O([[\^j (x+a)]{} ]{}).$$$$. +[j ]{}\^[j+1]{} (x+a) +j d\_j\^[(x)]{}(a) +j\_[=0]{}\^[j-1]{} (-1)\^![[j-1]{} ]{} \^[j--1]{} a  \_[+2]{}(-a)}$$$$=\_[x ]{} (\_[k=0]{}\^x [[\^j (k+a)]{} ]{} -[[\^[j+1]{} (x+a)]{}]{})+\^[j+1]{} a + jd\_j(a) +j\_[=0]{}\^[j-1]{} (-1)\^![[j-1]{} ]{} \^[j--1]{} a  \_[+2]{}(-a)$$$$=\_j(a)+\^[j+1]{}a +jd\_j(a)+j\_[=0]{}\^[j-1]{} (-1)\^![[j-1]{} ]{} \^[j--1]{} a  \_[+2]{}(-a). (31)$$ Part (b) has been demonstrated.

Remark. When $a=1$, the only term remaining in the sum on the right side of Eq. (8) is $\ell=j-1$. By relation (16), we then see the proper reduction 