I was thinking about the following:
Denote $\pi(x)$ as the prime counting function such that: $$ \pi(x) = \#\text{ of prime numbers}\leq x $$ It is well known from the prime number theorem that $$ \pi(x) \sim \frac{x}{\ln x} $$ and $$ \pi(x) \sim \text{Li}(x),\quad \text{Li}(x)=\int_2^x\frac{1}{\ln t}\,dt. $$ Note the following: if $A(x) \sim B(x)\implies A(x)/B(x) = 1 \text{ as }x\to\infty$ and $A(x), B(x)\to\infty \text{ as }x\to\infty$.
If $C(x)$ is a function such that $n \leq C(x) \leq k$ then:
$$(A(x) + n)/B(x) \leq (A(x) + C(x))/B(x) \leq (A(x) + k)/B(x)$$
$$A(x)/B(x) + n/B(x) \leq A(x)/B(x) + C(x)/B(x) \leq A(x)/B(x) + k/B(x)$$
If x is taken to infinity:
$$1 + 0 \leq A(x)/B(x) + C(x)/B(x) \leq 1 + 0$$
$$\rightarrow A(x) + C(x) ~ B(x)$$
What interests me is that since we already know that:
$$\pi(x) \sim x/\ln(x)$$
and from above that $x/ln(x) +$ any number of functions of the form $A(b(x)) ~ \pi(x)$
Can we not try to do some sort of fourier analysis on the function:
$$\pi(x) - x/\ln(x) $$
or
$\pi(x) - \text{Li}(x)$?