Let ¥Ä(x) = log ¥Ã(x + 1) x (.1 < x .= 0); ¥Ä(0) = .:
For all n = 0; 1; 2; : : : and x > .1, we show that (.1)n¥Ä(n+1)(x) = (n + 1)! ¡ò 1 0 un+1(n + 2; xu + 1) du;
where denotes the Hurwitz zeta function. This representation implies that ¥Ä¡Ç is completely monotonic on (.1;¡Ä). This extends a result published in 1996 by Grab- ner, Tichy, and Zimmermann, who proved that ¥Ä is increasing and concave on (.1;¡Ä).
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