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.

up vote 9 down vote accepted

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.

  • NIce! What Karp and Prilepkina do (in their Lemma 1) is differentiate the Dirichlet formula I was trying to use, to get a nicer integral expression for $\psi'$. Then the rest is indeed easy. – Iosif Pinelis Sep 1 '17 at 1:28
  • Nice, thanks! Positivity of $g(u)$ is easy to show. – MCH Sep 1 '17 at 2:17
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    @IosifPinelis I also started with differentiating the $\psi$ formula to get nice expressions, but then remembered that D. Karp had once sent me a paper with CM of Gamma functions, so decided to look into it for a more general result. Curiously, however, in general verifying $P(u)\ge 0$ is hard. Karp and Prilepkina provide some sufficient conditions also, and some necessary ones, but otherwise, seems not so easy to use their condition in general. – Suvrit Sep 1 '17 at 2:17
  • @Suvrit : I of course agree that in general it could be hard to verify $P(u)\ge0$. – Iosif Pinelis Sep 1 '17 at 3:00

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.

  • Beautiful. Thank you! – MCH Aug 31 '17 at 22:21
  • I suppose one can also directly use Euler's integral representation of Harmonic numbers: $H_n=\int_0^1 (1-x^n) dx/(1-x)$, which is essentially equivalent to your proof. – MCH Aug 31 '17 at 22:29
  • I am very sorry -- but the term $-v^x$ in the parentheses under the last integral sign was missing. – Iosif Pinelis Aug 31 '17 at 22:42
  • There should also be a multiplicative $\log v$ factor resulting from the derivative of $v^x$, right? I think that's what makes the derivative negative. The idea is clear, though. Thanks again! – MCH Aug 31 '17 at 22:53
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    $f'>0$ also follows because $v^{px}>v^x$ and $v^{qx}>v^x$ for $x>0$ and $p$ and $v$ in $(0,1)$. – Iosif Pinelis Sep 1 '17 at 0:13

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 $x \geq \tanh x$. $\square$

  • Thanks for filling in the missing details! Indeed, convexity here suffices to prove the desired superadditivity $g(p+q)\ge g(p)+g(q)$. – Suvrit Sep 1 '17 at 2:14
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    Another way to see positivity of $g$ is to prove $p/(e^u-1)\geq 1/(e^{u/p}-1)$, which is immediate by rearranging and taking the derivative in $u$. – MCH Sep 1 '17 at 2:20
  • @MahdiCheraghchi Oh, that's much better. Maybe the nicest way to see it is, since $e^x$ is convex, $\tfrac{e^{u/p}-1}{u/p} \geq \tfrac{e^u-1}{u}$. – David E Speyer Sep 1 '17 at 11:30

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