Proof of complete monotonicity of a binomial function

Question

By plotting the function and its derivatives, one can easily be convinced that the function $$f(x):=\log\binom{x}{p x}=\log\Gamma(x+1)-\log\Gamma(px+1)-\log\Gamma((1-p)x+1),$$ defined for $x>0$ and $p \in (0,1)$, is completely monotone (i.e., for all $x$, $f(x)>0$, $f'(x)<0$, $f''(x)>0$, etc). How can this statement be proved, or is there a known proof?

By Bernstein's theorem on monotone functions, the statement is equivalent to $f$ being the Laplace transform of a non-negative function. I tried to use the Taylor expansion of log-gamma $$\log\Gamma(x+1) = -\gamma x+\sum_{j=2}^\infty \frac{\zeta(j)}{j} (-x)^j,$$ and divide the $j$th term by $j!$ (which is basically what the inverse Laplace transform does). But the result doesn't seem to be a known function (without the $\zeta(j)$ constants, though, the resulting series would essentially become the expansion of the exponential integral function.

Edit: The fact that $f'(x)<0$ is equivalent to concavity of the digamma function, which can be shown by considering its integral form (see the first answer below). However, the higher derivatives seem less obvious to work with.

Answer 1

The following is the completely monotonic claim that actually holds (also hinted by Iosif Pinelis).

Claim. Let $f(x)=\log\binom{x}{px}$; then, $f''$ is CM.

We prove this claim as a corollary of the following impressive generalization.

Theorem. (Karp and Prilepkina, 2015). Consider the ratio \begin{equation*} W(x) := \frac{\prod_{i=1}^p \Gamma(A_ix+a_i)}{\prod_{j=1}^q\Gamma(B_jx+b_j)}, \end{equation*} where $(A_i,a_i)$ and $(B_j,b_j)$ are positive scalars. Then, $(\log W(x))''$ is CM if and only if \begin{equation*} P(u) = \sum_{i=1}^p \frac{e^{-a_iu/A_i}}{1-e^{-u/A_i}} - \sum_{j=1}^q \frac{e^{-b_ju/B_j}}{1-e^{-u/B_j}} \ge 0,\quad\forall u > 0. \end{equation*} In the affirmative case, $(\log W(x))'' = \int_0^\infty e^{-xu}uP(u)du$.

Proof of the claim. Let $q:= 1-p$. We need to check positivity of \begin{equation*} k(u) := \frac{e^{-u}}{1-e^{-u}} - \frac{e^{-u/p}}{1-e^{-u/p}} - \frac{e^{-u/q}}{1-e^{-u/q}}. \end{equation*} Observe that for $p\to 0$ or $p\to 1$, $k(u)\to 0$, which is promising. Simplifying, we reduce $k\ge 0$ to showing positivity of \begin{equation*} g(u) := \frac{1}{e^u-1} - \frac{1}{e^{u/p}-1} - \frac{1}{e^{u/q}-1}, \end{equation*} which I leave to the OP to verify.

Answer 2

This is a not a complete answer, but possibly a first step in it.

Using the Dirichlet formula (Theorem~1.6.1 in G. E. ANDREWS, R. ASKEY, AND R. ROY, Special functions, volume 71 of Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 1999) and then the substitution $v:=\frac1{1+z}$, one has \begin{equation} \psi(x+h)-\psi(x)=\int_0^\infty\Big(\frac1{(1+z)^x}-\frac1{(1+z)^{x+h}}\Big)\frac{d z}z =\int_0^1 v^{x-1}\frac{1-v^h}{1-v}d v \end{equation} for positive $x$ and $h$. So, \begin{equation} f'(x)=\int_0^1(pv^{px}+qv^{qx}-v^x)\frac{dv}{1-v}, \end{equation} where $q:=1-p$. In particular, it follows that $f'$ is positive (rather than negative). However, it appears that $f''$ is indeed completely monotone.

Answer 3

This is a footnote to Suvrit's answer: He reduces the problem to verifying the condition $$\frac{1}{e^u-1} \geq \frac{1}{e^{u/p}-1} + \frac{1}{e^{u/q}-1}$$ for $u>0$, $p+q=1$ and $p,q > 0$. Put $g(p) = \tfrac{1}{e^{u/p}-1}$ and extend continuously to $g(0)=0$. It is sufficient to show that $g$ is convex on $[0,1]$.

We have $$\frac{d^2 g}{(dp)^2} = \frac{e^{u/p} u}{p^3 (e^{u/p}-1)^2} \left( \frac{u}{p} \frac{e^{u/p}+1}{e^{u/p}-1}-2 \right)$$ so we just need to show the quantity in parentheses is positive. Putting $v=u/p$, we need to show $\tfrac{v (e^v+1)}{e^v-1} \geq 2$. We can rewrite this as $\tfrac{v}{2} \geq \tfrac{e^{v/2}-e^{-v/2}}{e^{v/2}+e^{-v/2}}= \tanh \tfrac{v}{2}$. Since the derivative of $\tanh x$ is $\tfrac{1}{\cosh^2 x}<1$, we have $x \geq \tanh x$. $\square$