\over {(s-1)^2}}
+O(x^{-s} \ln x), ~~~~~~~~~~ x \to \infty.  \eqno(98)$$

Additionally we may further generalize Proposition 3 to [**Proposition 4**]{}. We have for Re $a>0$, $j >-1$, and $|t| < 1$ (a) $$\sum_{k=2}^\infty (-1)^k \zeta(k,a) {t^{k+j} \over {k+j}}=\int_0^t
x^j[\psi(x+a)-\psi(a)]dx, \eqno(99)$$ and (b) $$(-1)^m\sum_{k=2}^\infty (-1)^k (k)_m\zeta(k+m,a) {t^{k+j} \over {k+j}}
=\int_0^tx^j[\psi^{(m)}(x+a)-\psi^{(m)}(a)]dx, \eqno(100)$$ or, equivalently, (c) $$\sum_{k=2}^\infty (-1)^k (k)_m\zeta(k+m,a) {t^{k+j} \over {k+j}}
=m!\int_0^tx^j[\zeta(m+1,x+a)-\zeta(m+1,a)]dx. \eqno(101)$$

Part (c) follows from part (b) by the use of relation (96). For the proof of part (a) we start with the identity $$\sum_{k=2}^\infty (-1)^k \zeta(k,a) x^{k-1}=\psi(x+a)-\psi(a).  \eqno(102)$$ We then multiply both sides of this equation by $x^j$ and integrate. The condition on $j$ is imposed due to the fact that $\psi(x)=-1/x-\gamma
+O(x)$ as $x \to 0$. For part (b) we repeatedly differentiate Eq. (75) with respect to $a$, use $${\partial^m \over {\partial a^m}}\zeta(k,a)=(-1)^m (k)_m \zeta(k+m,a),
\eqno(103)$$ and the Proposition is completed.

Remark. Proposition 4 may be extended by multiplying Eq. (102) by other functions and integrating over other intervals. For instance we have $$\sum_{k=2}^\infty (-1)^k \zeta(k,a)\int_0^t x^{k+\beta-1}(1-x)^\gamma
(1-zx)^{-\alpha}dx=\int_0^t x^\beta (1-x)^\gamma (1-zx)^{-\alpha}[\psi(x+a)
-\psi(a)]dx.  \eqno(104)$$ Once again such a result may be differentiated with respect to $a$ and Eqs. (103) and (96) used in order to obtain certain integrals of 