2005
DOI: 10.1016/j.jat.2004.02.008
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On a conjecture of Clark and Ismail

Abstract: Let m (x) = −x m (m) (x), where denotes the logarithmic derivative of Euler's gamma function. Clark and Ismail prove in a recently published article that if m ∈ {1, 2, . . . , 16}, then (m) m is completely monotonic on (0, ∞), and they conjecture that this is true for all natural numbers m. We disprove this conjecture by showing that there exists an integer m 0 such that for all m m 0 the function (m) m is not completely monotonic on (0, ∞).

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Cited by 13 publications
(27 citation statements)
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“…Completely monotonic functions attracted the attention of many authors and a plethora of results on completely monotonic functions have been obtained, most of them concerning functions defined in terms of the gamma and polygamma functions. We refer the reader to [5], [3], [17] [18], [20], [23] for some interesting results, applications, bibliography and helpful information regarding such functions.…”
Section: Let γ(X) Be Eulermentioning
confidence: 99%
“…Completely monotonic functions attracted the attention of many authors and a plethora of results on completely monotonic functions have been obtained, most of them concerning functions defined in terms of the gamma and polygamma functions. We refer the reader to [5], [3], [17] [18], [20], [23] for some interesting results, applications, bibliography and helpful information regarding such functions.…”
Section: Let γ(X) Be Eulermentioning
confidence: 99%
“…In [4], it was furthermore conjectured that (−x m ψ (m) (x)) (m) would be completely monotonic for all m ≥ 1 but it has turned out not to be the case; see [1]. In this direction we show that the function…”
Section: Introductionmentioning
confidence: 64%
“…Therefore, we will have to bring in some other ideas in order to reach the value of Q(x) which is less than −π/2. In order to find these new ideas, we have looked at the proof of the estimate Q(x) = Ω ± ( ln ln(x)) given in [1] and [10]. This proof is based on a rather interesting argument, which is both constructive and non-constructive.…”
Section: Numerical Resultsmentioning
confidence: 99%
“…Of course, it still takes a lot of work to deduce (19), as we have to show that the sum over all n which do not divide K is not too large. See [1] and [10] for all the details.…”
Section: Numerical Resultsmentioning
confidence: 99%