\eqno(46c)$$ where the second line follows from the first by integration by parts and $\Gamma(x,y)$ is the incomplete Gamma function. By using the Maclaurin series for $\ln(1-z)$ in either integral of Eq. (46a) we find $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}} = \sum_{r=1}^\infty {1 \over r}
\ln\left(1+ {1 \over r}\right).  \eqno(46d)$$ In addition, we have by partial summation $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}}=-\sum_{n=1}^\infty H_n\left[
{{\ln(n+2)} \over {n+2}}-{{\ln(n+1)} \over {n+1}}\right], \eqno(46e)$$ where $H_n \equiv \sum_{k=1}^n 1/k$ are the harmonic numbers, and we next show how this expression may be alternatively obtained from the integral representation (46b). We use the generating function $$-\ln(1-z)=(1-z)\sum_{n=1}^\infty H_n z^n \eqno(47)$$ at $z=e^{-t}$. We substitute for $\ln(1-e^{-t})$ on the right side of Eq. (46b), interchange summation and integration, justified by the absolute convergence of the integral, and find that $$\int_0^\infty e^{-t} \ln t \ln(1-e^{-t}) dt=-\sum_{n=1}^\infty H_n\int_0^\infty
[e^{-(n+1)t}-e^{-(n+2)t}]\ln t ~dt \eqno(48a)$$ $$=\sum_{n=1}^\infty H_n\left[\gamma\left({1 \over {n+1}}-{1 \over {n+2}}\right)
+{{\ln(n+1)} \over {n+1}}-{{\ln(n+2)} \over {n+2}}\right]. \eqno(48b)$$ We note that simply $H_{n-1}=H_n-1/n$ and recover Eq. (46e).

Using the generating function (47) and the right-most integral of Eq. (46a) we find that $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}}=\sum_{n=1}^\infty H_n
\int_0^1 {{(1-x)^2 x^{n-1}} \over {\ln x}} dx
=\sum_{n=1}^\infty H_n[\ln n -2\ln(n+1)+\ln(n+2)].  \eqno(49)$$ The integral of Eq. (49) may be 