Are $\lim_{x \to \infty} \frac{\Psi(x)}{\pi(x)}=1$ and $\pi(x)\le\Psi(x)$ true statements?

Question

$$\Psi(x)= \int_2^x \sin(\frac{1}{2\ln(t)})+\sinh(\frac{1}{2\ln(t)})\mathscr{dt}. $$

Does$$ \lim_{x \to \infty} \frac{\Psi(x)}{\pi(x)}=1? $$

Where $\pi(x)$ is the prime counting function.

Since $$ \Psi(x)\approx \mathscr{li}(x), $$

I believe the conjecture to be true.

Is $$\pi(x)\le\Psi(x)?$$

Answer 1

If it is true that $\Psi(x)\approx \text{Li}(x)$ and we use that $\pi(x)\approx\frac{x}{\ln(x)}$ then we can say that: $$\lim_{x\to\infty}\frac{\Psi(x)}{\pi(x)}=\lim_{x\to\infty}\frac{\ln(x)\int_0^x\frac{dt}{\ln(t)}}{x}$$ And now use L'Hopitals rule. Failing this use Squeeze theorem?

Answer 2

Since $\sin x=x+o(x)$ and $\sinh x=x+o(x)$, $$\Psi (x)=\int^x_2 \frac1{\ln t}dt+\int^x_2 o(\ln t)dt=\operatorname{Li}(x)+o(\operatorname{Li}(x))$$

Hence, $$\frac{\Psi(x)}{\pi(x)}\sim\frac{\operatorname{Li}(x)}{\pi(x)}\sim 1$$

provided that $\operatorname{Li}(x)/\pi(x)\to1$.