Existence and uniqueness of a function generalizing a finite sum of powers of logarithms 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.

Examples: 


*

*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?
 A: 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.
A: (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 $a$.
