Consider
$$ T(f) = \lim_{t \to \infty} \sum_{i =1}^t f(i) - \int_1^t f(x) dx $$
Notice $ T(a + b) = T(a) + T(b) $ and $T(C a) = C T(a)$ when $C $ is a constant. Also $T(0)= 0$.
Let $T(d) = 0$. So $T(1/x + d) = \gamma $
Now i Know sophomore's dream , so that is 1 solution to $d$ that is nontrivial.
Also $T(x^{-a}) = \zeta(a) - \frac{1}{a-1}$.
I wonder , is there a solution or many such that
$T(v) = \gamma $
Such that $v$ is no asymptotic of $\ln $ ?
In particular a solution $v_2$ such that both the sum and integral converge ?? And another one $v_3 $ where both the sum and integral diverge.
Both existence and closed forms are considered .
Also of particular interest are those $T(g) $ solutions above where for $x ≥ 1$ we have ,$ g '(x) $ is always the same sign.
This is perhaps not quite in the spirit of the question, but if $f(x)$ is a solution, so is $$ g(x) = f(x)-f(\lfloor x \rfloor); $$ the sum is identically zero, the integral converges if and only if the limit exists, and therefore $g$ will not be asymptotic to $\log$.
As for a divergent solution, $T(c)$ is defined and equal to zero for constant $c$, so $T(g+c)=T(g)$ for any $c$, while the sum is equal to $tc$ and the integral is $-tc + \int_0^t g$, both of which diverge if $c \neq 0$.
As for solutions with a derivative of constant sign, consider $f(x)=r^x$. The sum is $$ \frac{r}{1-r}(1-r^t), $$ while the integral is $$ -\frac{r}{\log{r}}(1-r^{t-1}). $$ Supposing that $0<r<1$, taking the limit gives $$ \frac{r}{1-r} + \frac{r}{\log{r}}. $$ One can show that this is a positive increasing function that maps $(0,1) \to (0,1/2)$. Hence one can find many nontrivial convergent solutions this way: for $0<r<1$, $$ T(ar^x) = a\left(\frac{r}{1-r} + \frac{r}{\log{r}} \right), $$ and one can find $a,r$ so that the right-hand side is $\gamma$ provided that $a>2\gamma$, and any $a$ gives exactly one such $r$.