Eq. (36): $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}}=\int_1^\infty f(t) dt+{1 \over 2}
f(1)+\int_1^\infty P_1(t)f'(t)dt$$ $$={\pi^2 \over {12}}+{1 \over 2}\ln^2 2+{1 \over 4} \ln 2
+\int_1^\infty P_1(t)f'(t)dt.  \eqno(43)$$ Combining Eqs. (42) and (43) gives Eq. (10a). Very similarly, we have $$\sum_{j=1}^\infty {{u^{j-1}} \over {j!}}(jd_j)=\int_0^\infty (t+a)^{u-1} 
\ln\left({{[t]+a+1} \over {t+a+1}}\right )dt,$$ giving $$\sum_{j=1}^\infty {{(-1)^{j-1}} \over {j!}}(jd_j)=\sum_{k=1}^\infty g(k,a)
-\left(1+{1 \over a}\right)\ln(a+1)-\ln a.$$ Applying Euler-Maclaurin summation then gives Eq. (10b), completing part (d) and the Proposition.

Remarks. (i) Concerning approximate numerical values in part (d) we have $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}} \simeq 1.25774688694, ~~~~~~~~~
\int_1^\infty P_1(t)f'(t)dt \simeq 0.0233638, \eqno(44a)$$ and $$\sum_{j=1}^\infty {{(-1)^{j-1}} \over {j!}}(jc_j) \simeq -0.12816.  \eqno(44b)$$ (ii) The interesting sum of Eq. (44a) has a multitude of equivalent expressions, including a great variety of series and integral representations. By splitting the sum over even and odd contributions we have $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}} =\ln^2 2 +{1 \over 2}
\sum_{k=1}^\infty {1 \over {(2k+1)}}\left[{{\ln(k+1)} \over {k+1}}+{{\ln(2k+1)} 
\over k}\right].  \eqno(45)$$

By using an integral representation for $\ln z$, we have derived the representations $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}} =\int_0^\infty {{(e^{-t}-1)} 
\over t}\ln(1-e^{-t}) dt
=\int_0^1 {{(1-x)\ln(1-x)} \over {x \ln x}}dx\eqno(46a)$$ $$=\int_0^\infty e^{-t}[1+\ln(1-e^{-t})]\ln t ~dt=-\gamma+\int_0^\infty e^{-t} \ln t
\ln(1-e^{-t}) dt,  \eqno(46b)$$ $$=-\gamma+\int_0^1 e^{-t} \ln t\ln(1-e^{-t}) dt-\sum_{k=1}^\infty {1 \over k}
{{\Gamma(0,k+1)} \over {(k+1)}},  