Today I spent my afternoon trying to understand why the Hardy-Littlewood's Second Conjecture is considered as a problem of such a great difficulty. I got a fairly strange result, so I decided to post it here and wait for opinion. So sorry if I made any kind of terrible mistake or so. The conjecture itself states that:
$$\pi(x+y) - \pi(x) \le \pi(y).$$
Or, in the same way,
$$\frac{\pi(x+y)}{y(y^2-1)} - \frac{\pi(x)}{y(y^2-1)} \le \frac{\pi(y)}{y(y^2-1)}.$$
Then, integrating both parts from $2$ to $\infty$ over the variable $y$,
$$\int_2^\infty \frac{\pi(x+y)}{y(y^2-1)}dy - \pi(x) \frac{\log\big(\frac{4}{3}\big)}{2} < \frac{\log(\zeta(2))}{2}.$$
Integrating by parts in the first integral,
$$\pi(x+2)\frac{\log\big(\frac{4}{3}\big)}{2} - \int_2^\infty \frac{\log\big(1-\frac{1}{y^2}\big)}{2}\pi '(x+y) dy - \pi(x) \frac{\log\big(\frac{4}{3}\big)}{2} < \frac{\log(\zeta(2))}{2}.$$
Then
$$\log\bigg(\frac{4}{3}\bigg) \left(\pi(x+2) - \pi(x) \right) - \int_2^\infty \log\bigg(1-\frac{1}{y^2}\bigg)\pi '(x+y) dy < \log(\zeta(2)).$$
And, as the difference between both Prime Counting Functions will be at most 1,
$$-\int_2^\infty \log\bigg(1-\frac{1}{y^2}\bigg)\pi '(x+y) dy < \log(\zeta(2))-\log\bigg(\frac{4}{3}\bigg) $$
$$-\int_2^\infty \log\bigg(1-\frac{1}{y^2}\bigg)\pi '(x+y) dy < \log\bigg(\frac{\pi^2}{8}\bigg) $$
Which seems to hold true, since the integral seems to monotonically decrease for large $x$.
Where is the flaw?
Thank you!
