Let $\pi(x)$ denote the prime counting function. It is well known that $\log \zeta(s) = \int_{2}^{\infty} \dfrac{s\pi(x)}{x(x^s - x)} \mathrm d{x}$ where $\Re(s)\geq 2$.
Inserting $s=4$, we have
$\log \zeta(4) = \log \dfrac{\pi^4}{945} = \int_{2}^{\infty} \dfrac{4\pi(x)}{x^4 - x} \mathrm d{x}$
From such a formula, why can't we determine the exact value (or at least sharper bounds), for $\pi (x)$ ?
EDIT: Infact it seems from this, we can verify the inequality $\mid \pi(x) - Li(x) \mid \leq \sqrt x \log x$ for all $x$ such that $\pi (x) > Li(x)$. We sketch the argument as follows:
Inserting $s=2$, we have $\log \zeta(2) = \log \dfrac{\pi^2}{6} = \int_{2}^{\infty} \dfrac{2\pi(x)}{x^2- x} \mathrm d{x}$. Suppose on the contradiction that $x$ is the least counterexample such that $\pi (x) > Li(x) + \sqrt x \log x$. Observe that this would imply that
$\log \zeta(2) > \int_{2}^{\infty} \dfrac{2(Li(x) + \sqrt x) \log x}{x^3- x} \mathrm d{x}$.
By the ineqaulity $Li(x)>\dfrac{x}{\log x}$ for all $x\geq 2$, the right hand side is $>$ $\int_{2}^{\infty} \dfrac{2(x + \sqrt x (\log x)^2)}{(x^2- x)\log x} \mathrm d{x}$, and by Wolfram one quickly finds that this is equal to $2.4296\cdots$.
But we now have $\log \zeta (2) = \log \dfrac{\pi^2}{6} = 0.498 > 2.4296$, an absurdity ?
Your mistake starts with "suppose by contradiction that..."
The opposite of "something is true for all $x$" is NOT "something is false for all $x$".
Your statement should be $\exists x$ such that $\pi (x) > Li(x) + \sqrt x \log x$. But then, since this is not true for all $x$ you don't get the integral inequality anymore.