(109) we use the representation $$\zeta(s,a)={a^{-s} \over 2} +{a^{1-s} \over {s-1}}-s\int_0^\infty {{P_1(x)} \over {(x+a)^{s+1}}}dx, ~~~~~~~\mbox{Re} ~s > -1,  \eqno(112)$$ that is based upon Euler-Maclaurin summation. For the proof of Eq. (93) we use Hermite’s formula for the Hurwitz zeta function (e.g., [@watson] p. 270), $$\zeta(s,a)={a^{-s} \over 2}+{a^{1-s} \over {s-1}}+2\int_0^\infty {{\sin(s\tan^{-1}y/a)} \over {(y^2+a^2)^{s/2}}}{{dy} \over {(e^{2\pi y}-1)}}.
\eqno(113)$$

Similarly, we may use a different integral representation for $\zeta(k,a)$ to obtain Eq. (111). We however describe a second method based upon the relation (96), whereby we obtain $$\sum_{k=2}^\infty (-1)^k \zeta(k,a) {t^{k} \over {k+j}}=\sum_{k=2}^\infty
{t^k \over {(k+j)}}{1 \over {(k-1)!}}\psi^{(k-1)}(a).  \eqno(114)$$ We make use of a classical Binet formula for the digamma function [@sri]. Upon repeated differentiation, we find $$\psi^{(n)}(z)=(-1)^{n-1}{{(n-1)!} \over z^n}-{{(-1)^n n!} \over {2z^{n+1}}}
-(-1)^n \int_0^\infty\left[{1 \over {1-e^{-t}}}-{1 \over t}-{1 \over 2}
\right]t^n e^{-tz} dt, ~~~~\mbox{Re}~z > 0.  \eqno(115)$$ Based upon this expression, inserting $\psi^{(k-1)}(a)$ into Eq. (114) and again interchanging summation and integration we determine Eq. (111). The interchange is justified by the absolute convergence of the integral of Eq. (115) and Proposition 5 is completed.

Remarks. It is possible to re-express the summation on the right side of Eq. (110). It is also possible to insert Binet’s formula for $\psi$ and $\psi^{(m)}$ 