I hope to find a proof of the following conjecture:

$(1)$ For every $a>0$ there is a convex analytic function $f_a:\mathbb R^+\to\mathbb R$ such that:

  • $f(1)=0$ and
  • $\forall x>1,\ f_a(x)=f_a(x-1)+\ln^ax$ (thus, for $n\in\mathbb N,\ f_a(n)=\sum_{k=1}^n\ln^ak$).

$(2)$ For every $a>0$ such function $f_a$ is unique.


  • for $a=1$, the function is $f_1(x)=\ln\Gamma(x+1)$.
  • for $a=2$, the function is $f_2(x)=\gamma_1+\frac{\gamma^2}2-\frac{\pi^2}{24}-\frac{\ln^2(2\pi)}2-\zeta''(0,x+1)$.
    (where $\gamma_1$ denotes the first Stieltjes constant, and $\zeta''$ denotes the second derivative of the Hurwitz $\zeta$-function with respect to its first parameter)

If the conjecture is true, can we find an explicit form for $f_a(x)$, e.g. an integral representation in terms of known special functions?

  • 1
    $\begingroup$ Doesn't $$C\pm\zeta^{(a)}(0,x+a-1)$$ does the trick? $\endgroup$ – Jack D'Aurizio Dec 16 '14 at 12:43
  • $\begingroup$ Yes, for $a\in\mathbb N$ it does (although I haven't proved its uniqueness). But how to make sense of $\zeta^{(a)}$ for fractional $a$? $\endgroup$ – Vladimir Reshetnikov Dec 16 '14 at 16:41
  • $\begingroup$ Have you tried the usual approach for defining the fractional derivatives, i.e. a Fourier transform, followed by a multiplication by a suitable power of the variable, followed by an inverse Fourier transform? $\endgroup$ – Jack D'Aurizio Dec 16 '14 at 17:12
  • $\begingroup$ Yes, I tried, but I could not obtain any explicit form (or any form suitable for a numeric approximation) for the fractional derivative, neither I proved that it would yield the desired result (i.e. that all conditions in the problem statement would be satisfied). $\endgroup$ – Vladimir Reshetnikov Dec 16 '14 at 17:42
  • $\begingroup$ Actually, for $x\in\mathbb N$ and any $a>0$ the fractional derivative assumes correct values (a finine sum of powers of logarithms), so it satifies the first condition and the second condition for integer arguments. Can we prove that it actually satisfies the second conditions for all real $x>1$, and that it is a convex and analytic function? $\endgroup$ – Vladimir Reshetnikov Dec 16 '14 at 18:13

I will show that there is a unique $f_a$ which obeys $f(1)=0$, $f(x) = f(x-1) + \log^a x$ and is convex on $(e^{a-1}, \infty)$. The functional equation gives us a unique extension to $(1,\infty)$; I am not sure whether it is convex there. (I will write $\log^a x$ for $(\log x)^a$, as the original poster does.)

Construction Set $g(x) = \log^a x$. So $$g'(x) = a \frac{\log^{a-1} x}{x} \ \mbox{and}$$ $$g''(x) = \left( \frac{d}{dx} \right)^2 \log^a x = \frac{a (a-1 - \log x) \log^{a-2} x}{x^2}.$$ For $x \in \mathbb{C} \setminus (-\infty, -1)$, set $$h_2(x) = - \sum_{n=1}^{\infty} g''(x+n).$$ (If $a<2$, we also have to remove $x=0$, so we don't have $\log 1$ in the denominator.) We have $g''(x+n) = O((\log n)^{a-1}/n^2)$, so the sum converges, and does so uniformly on compact sets. So $h_2(x)$ is an analytic function. Moreover, for $x \in (e^{a-1}, \infty)$, we have $g''(x) <0$, so $h_2(x)>0$. By construction, we have $$h_2(x) - h_2(x-1)=g''(x).$$ Also, easy estimates give $h_2(x) = O(\log^a x/x)$ as $x \to \infty$.

Put $h_1(x) = \int_{t=1}^x h_2(t) dt$. Then $h_1(x) - h_1(x-1) = g'(x) + C$ for some $C$. Sending $x$ to $\infty$, we have $$\lim_{x \to \infty} h_1(x) - h_1(x-1) = \lim_{x \to \infty} \int_{t=x-1}^x O(\log^a x/x) dt = O(\log^a x/x)=0,$$ and $\lim_{x \to \infty} g'(x) =0$, so $C=0$.

Integrating again, put $h(x) = \int_{t=1}^x h_1(t) dt$. So $h(x) - h(x-1) = \log^a x + C$ for some $C$. This time, I couldn't figure out whether or not $C$ is the correct constant. But, if it isn't, that's okay: Set $f(x) = h(x) - C(x-1)$. Now the condition $f(1) = 0$ and $f(x)-f(x-1) = \log^a x$ are okay, and $f''(x) = h''(x) = h_2(x)$ which, as we observed, is $>0$ for $x>e^{a-1}$.

Uniqueness Suppose there was some other $C^2$ function $\tilde{f}(x) = f(x)+r(x)$ which met the required criteria. Then $r(x)-r(x-1)=0$, so $r(x)$ is periodic. Suppose for the sake of contradiction that $r$ is not constant. (If $r$ is constant, then the constant is zero, as $r(1)=0$.) Then $r''$ is a periodic function with average $0$, so $r''(y)<0$ for some $y$, and we then have $r''(y)=r''(y+1)=r''(y+2)=\cdots$. But then $\tilde{f}''(y+n) = f''(y+n) + r''(y+n) = O(\log^{a-1} n/n^2)+r''(y+n)>0$ for all $n$, contradiction that $r''(y+n)<0$.

Convexity for all $x$? What remains is the question: Is $f''(x)>0$ for all $x \in (1,\infty)$? Or, equivalently, is $$\sum_{n=1}^{\infty} g''(x+n)<0$$ for all $x \in (1, \infty)$? I expected that the answer would be "no", and I would just have to search for a little bit to find a counter-example to finish this answer. But, so far, numerical computations suggest the answer is always "yes". I think I could prove this by unenlightening bounds, but instead I'm going to go to bed and see if I think of a better strategy in the morning.

  • $\begingroup$ I am missing why $h(x) = \int_1^xh_1(t)dt\implies\ h(x) - h(x-1) = \log^ax + C.\ $ The best I could do was $\ h(x) - h(x-1) = \log^ax + h_1(x-1) + \lim_{M\to\infty}\left(\sum_{N=1}^Mg'(x-1+N) - \log^a(x+M)\right)$ $\endgroup$ – will Aug 4 '15 at 19:25
  • $\begingroup$ Never mind. I simplified to $\ h(x) - h(x-1) = \log^ax + \lim_{P\to\infty}\sum_{N=1}^Pg'(1+N) - \log^a(b+P).\ $ Perhaps these are similar to the Stieltjes constants? $\endgroup$ – will Aug 4 '15 at 20:00
  • $\begingroup$ I'll answer anyways: $h'(x) - h'(x-1) = h_1(x) - h_1'(x-1) = \frac{d}{dx}(\log^a x)$, so $h(x)-h(x-1)=\log^a x+C$. $\endgroup$ – David E Speyer Aug 4 '15 at 20:54
  • $\begingroup$ Have you been aware of this math.stackexchange.com/a/864073/1714 earlier question (and answers)? $\endgroup$ – Gottfried Helms Sep 8 '16 at 8:15
  • $\begingroup$ @GottfriedHelms Nope! Thanks for pointing it out. It doesn't look like either that answer or this one resolves the question of convexity on $[1,e^{a-1}]$, though? $\endgroup$ – David E Speyer Sep 8 '16 at 12:15

(The inclusion of $\ln^a(1+\epsilon)$ in the original definition is quite useful, but not very smooth)

Suppose we had a contour $\mu_x$ that descended from $+i\infty$, passing below the real axis at $z_1=1$, and passing up back towards $+i\infty$ at some $z_x\in\Re(x,x+1)$ Then we have residues, for $0 < \Re(a)$ and $x\in\Re(1,\infty)$: $$ f_a(x) = \oint_{z\in\mu_x}\frac{\ln^a(z)}{e^{2\pi i(z-x)}-1}dz = \sum_{k=0}^{\lfloor x-1\rfloor}\ln^a(x-k) $$ I denote the sum of the residues with $f_a,\ $ despite $f_a(x)$ being neither analytic nor convex in $x\in\Re_+$. $f_a$ is continuous and satisfies $f_a(1) = 0$ and $1 < x\implies\ f_a(x) - f_a(x-1) = \ln^a(x).\ $ Perhaps we can construct an explicit contour: $$ \mu_x(t) = \frac{2i}{1+t} + 1-3i + \frac{2x-1}{2}t + \frac{2i}{2-t},\ \ \ t\in\Re(-1,2) $$ $$ f_a(x) = \int_{-1}^2\frac{\ln^a\mu_x(t)}{e^{2\pi i(\mu_x(t)-x)}-1}\left( \frac{2x-1}{2} - \frac{2i}{(1+t)^2} + \frac{2i}{(2-t)^2} \right)dt $$ In the neighborhood of any given $0 < x$, and for $t\in\Re[0,1],\ $ we notice that the integrand and its $t$-derivative are bounded absolutely. So we can take: $$ \frac{d}{dx}f_a(x) = \int_{-1}^2\frac{x\ln^a(\mu_x(t))dt}{e^{2\pi i(\mu_x(t)-x)}-1} + \ldots $$$$ \int_{-1}^2\frac{d}{dx}\left(\frac{\ln^a\mu_x(t)} {e^{2\pi i(\mu_x(t)-z)}-1}\right)\left( \frac{2x-1}{2}-\frac{2i}{(1+t)^2} + \frac{2i}{(2-t)^2}\right)dt $$ $$ t\approx 0\implies\ln^a\mu_x(t)\approx\ln^a(1+t\frac{2x-1-3i}{2}) $$ $$ \mbox{(a somewhat problematic place for a derivative)} $$ When we are looking for convex functions that almost satisfy (1), we might consider the use of remainder operator, $R(x):= x -\lfloor x\rfloor$, and append our finite sum with: $$ f_a(x) = (R^a(x)-1)R^a(x) + \sum_{k=0}^{\lfloor x-1\rfloor}\ln^a(x-k) $$ Which still statisfies, $1 < x\implies\ f_a(x) - f_a(x-1) = \ln^a(x).\ $ In conclusion, I suspect we can find many such functions continuous and convex on $x\in\Re(1,\infty)$, but none that satisfy (1) for fractional $a$.

  • $\begingroup$ This got the bounty, so I assume it is a good answer, but I am puzzled. It seems like the function you construct is given by a parabola on each interval (n, n+1), and not analytic at all. Do I misunderstand the definition? $\endgroup$ – David E Speyer Aug 4 '15 at 4:14
  • $\begingroup$ At the end of my answer I incorrectly implied that I found analytic sums. I have edited this to continuous. I would claim that these finite sums are $\lfloor a \rfloor$ times differentiable. $\endgroup$ – will Aug 4 '15 at 16:49

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