{n \choose k}\left[{1 \over {(k-2)!}}\int_1^\infty {{P_1(x)} \over x^{m+1}}\ln^{k-2}
x ~dx\right.$$ $$\left.+{1 \over {(m-1)^k}}-{m \over {(k-1)!}}\int_1^\infty {{P_1(x)} \over x^{m+1}}
\ln^{k-1}x ~dx\right]$$ $$=\sum_{m=2}^\infty {{(-1)^m} \over m}\left[-\int_1^\infty {{P_1(x)} \over x^{m+1}}
L_{n-2}^2(\ln x)dx+1-{n \over {(m-1)}}-\left({{m-2} \over {m-1}}\right)^n \right.$$ $$\left.+ m\int_1^\infty {{P_1(x)} \over x^{m+1}}[n-L_{n-1}^1(\ln x)] dx \right]. 
\eqno(37)$$ We then make use of the elementary series $$\sum_{m=2}^\infty {{(-1)^m} \over m}=1-\ln 2, ~~~~~~~~~~~
-n\sum_{m=2}^\infty {{(-1)^m} \over {m(m-1)}}=n-2n \ln 2,  \eqno(38a)$$ $$\sum_{m=2}^\infty {{(-1)^m} \over m}{1 \over x^{m+1}}={1 \over x^2}-{1 \over x}
\ln\left(1+{1 \over x}\right), ~~~~~~~~~\sum_{m=2}^\infty {{(-1)^m} \over x^{m+1}}
={1 \over {x^2(1+x)}},  \eqno(38b)$$ and part (c) of Proposition 1 follows. For part (d) we may first form using Eq. (6) $$\sum_{j=1}^\infty {{u^{j-1}} \over {j!}}(jc_j)=\int_1^\infty t^{u-1} 
\ln\left({{[t]+1} \over {t+1}}\right )dt, ~~~~~~~~~~\mbox{Re}~u \leq 0.
\eqno(39)$$ We may note in passing that this equation properly recovers the expression for $c_1$ at $u=0$. At $u=-1$ we have $$\sum_{j=1}^\infty {{(-1)^{j-1}} \over {j!}}(jc_j)=\sum_{k=1}^\infty
\int_k^{k+1} {{\ln(k+1)} \over t^2} dt - \int_1^\infty {{\ln(t+1)} \over t^2}
dt.  \eqno(40)$$ An integration by parts gives the integral $$\int_1^x {{\ln(t+1)} \over t^2}dt=\ln x -\left(1+{1 \over x}\right)\ln(x+1)
+2\ln 2.  \eqno(41)$$ Therefore we have from Eq. (40) $$\sum_{j=1}^\infty {{(-1)^{j-1}} \over {j!}}(jc_j)=\sum_{k=1}^\infty {{\ln(k+1)}
\over {k(k+1)}} - 2 \ln 2.  \eqno(42)$$ We next apply Euler-Maclaurin summation to the sum on the right side of 