\over 2}\ln^2 2-{7 \over 4}\ln 2+\int_1^\infty P_1(t)f'(t)dt,  \eqno(10a)$$ while with $g(t,a)=\ln(t+a)/(t+a-1)(t+a)$ $$\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$$ $$=\zeta(2)+{1 \over 2}\ln^2 (a+1)+\mbox{Li}_2(-a)+{1 \over 2}{{\ln(a+1)} \over
{a(a+1)}}-\left(1+{1 \over a}\right)\ln(a+1)-\ln a+\int_1^\infty P_1(t)g'(t,a)dt.
\eqno(10b)$$ In Eq. (8) Li$_j$ is the polylogarithm function and in Eq. (9) $\gamma$ is the Euler constant.

After the proof of this Proposition we present additional summatory results and a generalization of them. These other Propositions pertain to $\gamma_0(a)$ and sums of Hurwitz zeta function values. The final section and Proposition treat the coefficients of the Laurent expansion of the logarithmic derivative of the Riemann zeta function about $s=1$.

**Proof of Proposition 1**

For part (a) we begin by forming $$\sum_{m=2}^\infty {{(-1)^m} \over m}(-1)^j \zeta^{(j)}(m)
=\sum_{k=2}^\infty \sum_{m=2}^\infty {{(-1)^m} \over {mk^m}} \ln^j k$$ $$=\sum_{k=2}^\infty \ln^j k\left[{1 \over k} -\ln\left({{k+1} \over k}\right)
\right]$$ $$=\lim_{x \to \infty}\left \{\sum_{k=2}^x {{\ln^j k} \over k}+\sum_{k=2}^x
[\ln k-\ln(k+1)]\ln^j k\right \}.  \eqno(11)$$ We then apply summation by parts to write $$\sum_{k=2}^x [\ln k-\ln(k+1)]\ln^j k=-\ln([x]+1)\ln^j x+j\int_1^x \ln([t]+1)
{{\ln^{j-1} t} \over t}dt$$ $$=-\ln([x]+1)\ln^j x+j\left(\int_1^x \ln(t+1){{\ln^{j-1} t} \over t}dt+c_j^{(x)}
\right),  \eqno(12)$$ where $$c_j^{(x)} \equiv \int_1^x {{\ln^{j-1} t} \over t} \ln\left({{[t]+1} \over
{t+1}}\right )dt, ~~~~~~~~~~j \geq 1.  \eqno(13)$$ The integral on the right side of Eq. (12) is evaluated 