Prove an inequality for the cosine integral

Question

For$$f(x)=\operatorname{Ci}(x)-\ln\left(1-\frac1x\right),x>0$$ I want to prove $$f(x)>0$$and $$f(x+2\pi)<f(x)$$ where $\operatorname{Ci}(x)=\int_x^\infty{\frac{\cos(t)}tdt}$.
Consider the integral form of $\operatorname{Ci}$, I turn $\ln(1-\frac1x)$ into $$\int_x^\infty{\frac1{t(1-t)}dt}$$(but it seems to be useless after I tried a lot)
For the first one, I think these 2 steps are needed:
(1):Prove $$f(x)>0,x=\left(2k+\frac32\right)\pi,k\in\mathbb{N}$$ (2):Base on (1), Prove$$f(x-t)>0,x=\left(2k+\frac32\right)\pi,k\in\mathbb{N},0<t<2\pi$$ But I am stuck here.
I have no idea about the second one but integrate $\frac{\cos(t)}t-\frac1{t(1-t)}dt$. Obviously, it doesn't work.

Answer 1

Assuming you want to prove such inequality for $x>1$, by the (inverse) Laplace transform we have

$$\text{Ci}(x)=-\int_{x}^{+\infty}\frac{\cos(z)}{z}\,dz=\int_{0}^{+\infty}\frac{\sin(x)-s\cos(x)}{s^2+1}e^{-sx}\,ds $$ $$ \log\left(\frac{x}{x-1}\right) = \int_{0}^{+\infty}\frac{e^s-1}{s}e^{-sx}\,ds $$ hence it is enough to notice that $-\frac{1}{\sqrt{s^2+1}}+\frac{e^s-1}{s}$ is positive on $\mathbb{R}^+$, where $\left|\sin(x)-s\cos(x)\right|\leq\sqrt{s^2+1}$ is a consequence of the Cauchy-Schwarz inequality.