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,

$$\pi(x+2)\frac{\log\big(\frac{4}{3}\big)}{2} - \int_2^\infty \frac{\log\big(1-\frac{1}{y^2}\big)}{2}\pi '(x+y) dy - \pi(x) \frac{\log\big(\frac{4}{3}\big)}{2} < \frac{\log(\zeta(2))}{2}.$$


$$\log\bigg(\frac{4}{3}\bigg) \left(\pi(x+2) - \pi(x) \right) - \int_2^\infty \log\bigg(1-\frac{1}{y^2}\bigg)\pi '(x+y) dy < \log(\zeta(2)).$$

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!

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    $\begingroup$ The flaw is that you start with the conjectore $A$ adn derive some other statement $B$ and notice that $B$ is true. However, from $A\implies B$ and $B$ we cannot infer $A$. $\endgroup$ – Hagen von Eitzen Feb 8 '17 at 18:26
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    $\begingroup$ No, @HagenvonEitzen just notes that you take the expression that you want to prove, manipulate it, giving something that is true, and hope to conclude from that that the conjecture is true. This is not a logically correct argument though. $\endgroup$ – user2520938 Feb 8 '17 at 18:31
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    $\begingroup$ The fact that the integral of one side is larger then the integral if the other side, doesn't mean that the one side is pointwise smaller then the other side $\endgroup$ – user2520938 Feb 8 '17 at 18:36
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    $\begingroup$ $\forall y\colon f(y)<g(y)$ implies, but is not equivalent to $\int f(y)\,\mathrm dy<\int g(y)\,\mathrm dy$. $\endgroup$ – Hagen von Eitzen Feb 8 '17 at 19:05
  • 1
    $\begingroup$ Nice! Thank you very much for pointing out my mistake $\endgroup$ – user3141592 Feb 8 '17 at 19:22

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