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?

(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 $a$.
