here is $$\sum_{q=0}^\infty \ln^\ell(q+k/m)\left[{1 \over {mq+k}}-\ln\left({{mq+k+1} \over
{mq+k}}\right)\right]$$ $$=\lim_{x \to \infty}\left \{\sum_{q=0}^x {{\ln^\ell(q+k/m)} \over {mq+k}} +\sum_{q=0}^x [\ln (mq+k)-\ln(mq+k+1)]\ln^\ell (q+k/m)\right \}.  \eqno(63)$$ Summing by parts, we have $$\sum_{q=0}^x [\ln (mq+k)-\ln(mq+k+1)]\ln^\ell (q+k/m)=[\ln k-\ln(k+1)]\ln^\ell (k/m)$$ $$+B(x)\ln^\ell (x+k/m)-\ell\int_0^xB(t){{\ln^{\ell-1} (t+k/m)} \over {t+k/m}}dt,  
\eqno(64)$$ where we have put $$B(x) = \sum_{q=1}^x [\ln(mq+k)-\ln(mq+k+1)].  \eqno(65)$$ The integral term of Eq. (64) gives an extension of the integral of Eq. (26) when $m>1$. From Proposition 1(b) we obtain [**Corollary 1**]{}. We have for Re $a>0$ and Re $b>0$ $$\ln\left[{{\Gamma(b)} \over {\Gamma(a)}}\right]=b\ln^2 b-a\ln^2a+\int_0^\infty
[1+\ln(a+t)]\ln\left({{[t]+a+1} \over {t+a+1}}\right)dt-\int_0^\infty
[1+\ln(b+t)]\ln\left({{[t]+b+1} \over {t+b+1}}\right)dt$$ $$+\sum_{k=0}^\infty \ln(k+a)\left[1-(k+a)\ln\left({{k+a+1}\over {k+a}}\right)\right]
-\sum_{k=0}^\infty \ln(k+b)\left[1-(k+b)\ln\left({{k+b+1}\over {k+b}}\right)\right]$$ $$+b+\ln b[-b+(1+b)\ln(b+1)]+\mbox{Li}_2(-b)
-a+\ln a[a-(1+a)\ln(a+1)]-\mbox{Li}_2(-a).  \eqno(66)$$

In contrast, a standard sum representation is [@grad] (p. 936) $$\ln\left[{{\Gamma(b)} \over {\Gamma(a)}}\right]=\ln\left({a \over b}\right)
+\sum_{k=1}^\infty \left[(b-a)\ln\left(1+{1 \over k}\right)+\ln\left({{k+a} 
\over {k+b}}\right)\right].   \eqno(67)$$

Proof. We use the summatory relation [@coffeystdiffs] (Proposition 1) $$\sum_{n=0}^\infty {1 \over {n!}}[\gamma_{n+1}(a)-\gamma_{n+1}(b)]=\ln\left[
{{\Gamma(b)} \over {\Gamma(a)}}\right], ~~~~\mbox{Re} ~a>0, ~~\mbox{Re} ~b>0. 
\eqno(68)$$ We write $$(-1)^j\sum_{m=2}^\infty {{(-1)^m} \over m}\left[\zeta^{(j)}(m,a)-\zeta^{(j)}(m,b)
\right]=\sum_{k=0}^\infty \ln^j(k+a)\left[{1 \over {k+a}}-\ln\left({{k+a+1} \over
{k+a}}\right)\right]$$ $$-\sum_{k=0}^\infty \ln^j(k+b)\left[{1 \over {k+b}}-\ln\left({{k+b+1} \over
{k+b}}\right)\right].  \eqno(69)$$ Putting $j=n+1$ and summing $\sum_{n=0}^\infty (1/n!)$ gives the terms of line $2$ of the Corollary. The integral terms of the Corollary result from using the definition (5) of $d_{n+1}(y)$.

For the polylogarithm terms we have $$\sum_{n=0}^\infty {{(n+1)} \over 