differences of Hurwitz zeta functions. Moreover, Eq. (100) could again be multiplied by say $t^n$ and integrated. For special values of $t$ including $1$ explicit results are obtainable by using various representations of the digamma function. By using various integral representations of the Hurwitz zeta function we may determine evaluations for the integrals of Proposition 4. We illustrate this in relation to Eq. (92). If we put for integers $j \geq 0$, $k \in C$, and $|t|<1$, $$M_j(k,t,a) \equiv \sum_{m=2}^\infty {{(-1)^m} \over {(m+j)}}{t^m \over {(k+a)^m}},
\eqno(105)$$ then we have $$M_j(k,t,a)={1 \over {(j+2)}}{t^2 \over {(k+a)^2}} ~_2F_1\left(1,j+2;j+3;-{t \over
{k+a}}\right) \eqno(106)$$ $$=(-1)^j t^{-j}(k+a)^j\left[M_0(k,t,a)-\sum_{m=2}^{j+1}{{(-1)^m t^m}
\over {m(k+a)^m}}\right], \eqno(107)$$ and $$M_0(k,t,a)={t \over {k+a}}-\ln\left(1+{t \over {k+a}}\right).  \eqno(108)$$ These expressions follow in like manner to Lemma 4. We then have [**Proposition 5**]{}. For Re $a>0$, integers $j \geq 0$, and $|t| < 1$ we have $$\sum_{k=2}^\infty (-1)^k \zeta(k,a) {t^{k} \over {k+j}}=\left({1 \over 2}-
{a \over {(j+1)}}\right)M_j(0,t,a)+{t \over {j+1}}\ln(1+t/a)$$ $$-\int_0^\infty\left[{t^2 \over
{(x+a)^2}}{1 \over {(x+t+a)}}-jM_j(x,t,a)\right]P_1(x)dx \eqno(109)$$ $$=\left({1 \over 2}-{a \over {(j+1)}}\right)M_j(0,t,a)+{t \over {j+1}}\ln(1+t/a)
-2\int_0^\infty \sum_{k=2}^\infty {{\sin(k \tan^{-1} y/a)} \over {(a^2+y^2)^{k/2}}}
{{dy} \over {(e^{2\pi y}-1)}}  \eqno(110)$$ $$=\left({1 \over 2}-{a \over {(j+1)}}\right)M_j(0,t,a)+{t \over {j+1}}\ln(1+t/a)$$ $$+\int_0^\infty \left({1 \over {1-e^{-w}}}-{1 \over w}-{1 \over 2}\right){e^{-aw}
\over w} M_j(0,t,1/w)dw.  \eqno(111)$$

In the proof of Eq. 