Asymptotic Expression for the Twin Prime Counting Function A variation on a previous question I asked, which has garnered no responses. I'll attempt to be more lucid:
Let $\pi_2(x)$ be the twin prime counting function and $\pi(x)$ be the prime counting function. An explicit formula would be:
$\pi_2(x) = \sum_{n = 3}^{x}[\pi(n) - \pi(n-1)][\pi(n+2) - \pi(n)]$
Basically, the expression inside the sum will equal $1$ when $n$ is the first item of a twin prime pair, and $0$ otherwise. We can expand it to:
$\pi_2(x) = \sum_{n = 3}^{x}\pi(n)\pi(n+2) - \pi(n)^2 - \pi(n-1) \pi(n+2) + \pi(n)\pi(n-1)$
and since $\pi(n)$ is a discrete counting function:
$\pi_2(x) = \int_{n = 3}^{x}\pi(n)\pi(n+2) - \pi(n)^2 - \pi(n-1) \pi(n+2) + \pi(n)\pi(n-1)$
From here, are we allowed to use the prime number theorem ($\pi(x) \sim \frac{n}{ln(n)}$) to get an asymptotic expression for $\pi_2(x)$? Namely, does:
$\pi_2(x) \sim \int_{n = 3}^{x}\frac{n}{ln(n)}\frac{n+2}{ln(n+2)} - (\frac{n}{ln(n)})^2 - \frac{n-1}{ln(n-1)} \frac{n+2}{ln(n+2)} + \frac{n}{ln(n)}\frac{n-1}{ln(n-1)}$ ?

$$
\pi_2(x) = \int_{n = 3}^{x}\left(\pi(n)\pi(n+2) - \pi(n)^2 - \pi(n-1) \pi(n+2) + \pi(n)\pi(n-1)\right)dn
$$
and
$$
\pi(n)\sim\frac{n}{\log n}
$$
means that
$$
\pi(n)=(1+o(1))\frac{n}{\log n}
$$
and so
$$
\pi_2(x) = \int_{n = 3}^{x}\left(
(1+o(1))\frac{n}{\log n}\frac{n+2}{\log(n+2)} - (1+o(1))\frac{n^2}{\log^2n} - 
(1+o(1))\frac{n-1}{\log(n-1)}\frac{n+2}{\log(n+2)} + (1+o(1))\frac{n}{\log n}\frac{n-1}{\log(n-1)}
\right)dn
$$
and hence
$$
\pi_2(x) = \int_{n = 3}^{x}\left(
\frac{n}{\log n}\frac{n+2}{\log(n+2)} - \frac{n^2}{\log^2n} - 
\frac{n-1}{\log(n-1)}\frac{n+2}{\log(n+2)} + \frac{n}{\log n}\frac{n-1}{\log(n-1)}+o\left(\frac{n^2}{\log^2n}\right)
\right)dn
$$
which is dominated by its error term
$$
\pi_2(x) = \int_{n = 3}^{x}o\left(\frac{n^2}{\log^2n}\right)dn
$$
and hence
$$
\pi_2(x) = o\left(\frac{x^3}{\log^2x}\right).
$$
Moral of the story: error terms are very important, especially in analytic number theory.
