$\def\d{\mathrm{d}}$After I was studying variations of the integral representation for $\zeta(3)$ due to Beukers, see the section More complicated formulas from this Wikipedia, I've thought an exercise. The motivation is this antiderivative $$-\int_0^1e^y\frac{\log(xy)}{1-xy}\,\d x=-\frac{e^y}{y}\operatorname{Li}_2(1-xy)+\mathrm{constant}.$$ For real numbers $x\geq 2$, $$f(x):=-\int_0^1\frac{e^y}{y}(\operatorname{Li}_x(1-y)-\zeta(x)) \,\d y,$$ I've calculated with Wolfram Alpha online calculator some particular values at integer values $x=n\geq 2$.
Question.
A) Is it possible to prove that $f(x)$ is decreasing for $x\geq 2$?
B) Is it possible to calculate $$\lim_{x\to\infty}f(x)?$$ Thanks in advance.
Also I know Euler's Zeta function $\zeta(x)$ tends to $1$ as $x\to\infty$. With respect the definite integral I believe that is (impossible) difficult to get a closed-form for such values.
Answer for question $(A)$ :
$\displaystyle -\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy =\sum\limits_{k=1}^\infty\frac{1}{k^x}\int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy$
With $\enspace\displaystyle 0\leq (e^y-1)\frac{1-(1-y)^k}{y}-y<1\enspace$ for $\enspace 0\leq y\leq 1$
(see Value range for $a$ when $\enspace 0\leq (e^x-1)\frac{1-(1-x)^a}{x}-x<1\enspace$ with $\enspace 0\leq x\leq 1$ .) we get
$\displaystyle 0<y+\frac{1-(1-y)^k}{y}\leq e^y\frac{1-(1-y)^k}{y}<1+y+\frac{1-(1-y)^k}{y}\enspace$ and therefore
$\displaystyle \int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy<\int\limits_0^1 (1+y+\frac{1-(1-y)^k}{y})dy=\frac{3}{2}+H_k\enspace$ with $\enspace\displaystyle H_k:=\sum\limits_{v=1}^k\frac{1}{v}$ .
It follows $\enspace\displaystyle 0<-\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy<\sum\limits_{k=1}^\infty\frac{1}{k^x}(\frac{3}{2}+H_k)=\frac{3}{2}\zeta(x)+\sum\limits_{k=1}^\infty\frac{H_k}{k^x}$ .
With $\enspace\gamma\enspace$ as the Euler–Mascheroni constant and $\enspace H_{k-1}<\gamma+\ln k\enspace$ for $\enspace k\in\mathbb{N}\enspace$ follows
$\displaystyle 0<-\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy<(\frac{3}{2}+\gamma)\zeta(x)+\zeta(x+1)-\zeta'(x)$
so that the integral convergies for $\enspace x>1$ .
The integral is decreasing because of $\enspace\displaystyle \frac{1}{k^a}>\frac{1}{k^b}\enspace$ for $\enspace k>1\enspace$ and $\enspace 0<a<b$ ,
together with $\enspace\displaystyle\int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy>0\enspace$ for all $\enspace k$ .
Answer for question (B):
$\displaystyle \lim\limits_{x\to\infty}(-\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy) =\lim\limits_{x\to\infty}(\sum\limits_{k=1}^\infty\frac{1}{k^x}\int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy)=$
$\displaystyle =\int\limits_0^1 e^y\frac{1-(1-y)^1}{y}dy=e-1$