I am wondering if the Gamma function is completely monotone within the interval (0,1). I use Mathematica to plot the derivatives of Gamma function up to 10th order of derivatives and the results looks like it is completely monotone. Is it possible to prove it mathematically?
Yes! Recall that the derivative of a gamma function can be expressed using the di-gamma function, Γ′ = Γψ, and for di-gamma ψ∈(0,1) the di-gamma is negative while Γ>0 for ∈(0,1) which satisfies monotonicity. As you state, higher-order derivates of Γ can be expressed as "poly-gamma" functions ψn with a sign alternation that also satisfies monotonicity. It's gonna take too long to write this out here to prove this to you but the textbook Explorations in Complex Functions delineates this in detail
Another proof without computations : Since Gamma is logarithmic convex on
(0,+infty), and takes the same values at 1 and 2, it has to be deceasing on (0,1).
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