of part (b) to part (a).

For the proof of part (c) we make repeated use of [**Lemma 2**]{}. We have $$\sum_{j=\nu}^n {{(-1)^{j-1}} \over {(j-\nu)!}}{n \choose j} w^{j-\nu}
=(-1)^{\nu-1} L_{n-\nu}^\nu (w),  \eqno(32)$$ that follows from the power series form of the Laguerre polynomials. We also apply [**Lemma 3**]{}. We have for integers $n \geq 1$ $$\zeta^{(n)}(s)=(-1)^n n\int_1^\infty {{P_1(x)} \over x^{s+1}}\ln^{n-1} x ~dx
-(-1)^n s\int_1^\infty {{P_1(x)} \over x^{s+1}}\ln^n x ~dx +{{(-1)^n n!} \over
{(s-1)^{n+1}}}.  \eqno(33)$$ Lemma 3 follows from induction on the integral representation [@titch] (p. 14) $$\zeta(s)={1 \over 2}+{1 \over {s-1}}-s\int_1^\infty {{P_1(x)} \over x^{s+1}}dx.
\eqno(34)$$ We combine the definition (3) of $S_\gamma(n)$, Lemma 2, the definition (6) of $c_j$, and the expression (7) for $\gamma_j$. For the $\zeta(j)$ values in Eq. (7) we apply the integral representation for Re $s>1$ and Re $a>0$ $$\Gamma(s)\zeta(s,a)=\int_0^\infty {{t^{s-1} e^{-(a-1)t}} \over 
{e^t-1}}dt,  \eqno(35)$$ at $a=1$. We then obtain $$S_\gamma(n)=-\gamma n-\sum_{k=2}^n {1 \over {(k-1)!}}{n \choose k}
\sum_{m=2}^\infty {{(-1)^m} \over m}\zeta^{(k-1)}(m)$$ $$+\int_0^\infty \left[L_{n-1}^1(-x)-L_{n-1}^1(-x/2)\right]{{dx}
\over {(e^x -1)}}+\int_1^\infty {{L_{n-2}^2(\ln t)} \over t}
\ln\left({{[t]+1} \over {t+1}}\right )dt.  \eqno(36)$$

We use Lemma 3 to re-express the double series on the right side of Eq. (36): $$\sum_{k=2}^n {1 \over {(k-1)!}}{n \choose k} \sum_{m=2}^\infty {{(-1)^m} \over m}
\zeta^{(k-1)}(m)=\sum_{m=2}^\infty {{(-1)^m} \over m}\sum_{k=2}^n (-1)^{k-1}
