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Someone deduced without using complex analysis that

$$ \int \frac{\pi(t)}{t^2} \mathrm{d}t \sim \log\log t $$ where $\pi$ is the prime counting function.

By differentiating the above, he then arrives at

$$\frac{\pi(t)}{t^2} \sim \frac{1}{t\log t} $$ which is exactly the Prime Number Theorem.

However, he feels that something should be wrong with this approach, but not sure exactly what ?

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    $\begingroup$ Where did you bring the first relation from? $\endgroup$
    – Arash
    Commented Nov 26, 2017 at 15:38
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    $\begingroup$ @ArashMohammadi It follows e.g. from one of Mertens' estimates for $\sum1/p$ and Abel's summation formula. $\endgroup$
    – Bart Michels
    Commented Nov 26, 2017 at 15:40
  • $\begingroup$ Supplemental to the answers below: If $f(t) \sim g(t)$ as $t\to \infty$ then $f(t)=g(t)(1+h(t))$ where $h(t)\to 0$ as $t\to \infty.$ So $f'(t)/g'(t)=1+h(t)+g(t)h'(t)/g'(t).$ Without more info about $h(t)$ we can't even estimate $h'(t),$ let alone estimate $f'(t)/g'(t).$ $\endgroup$
    – DanielWainfleet
    Commented Nov 28, 2017 at 0:53

2 Answers 2

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The reasoning is flawed because $f\sim g$ most certainly does NOT imply $f’\sim g��$.

For example, take $f(x)\equiv 0$ and $g(x)= \frac1N \sin N^2x$.

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    $\begingroup$ @barto it means $f/g\to 1$. $\endgroup$
    – Vim
    Commented Nov 26, 2017 at 16:44
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    $\begingroup$ To avoid the zeroes-problem, I'd suggest to take something like $f(x)=x$ and $g(x)=x+\sin x^2$ $\endgroup$
    – Hagen von Eitzen
    Commented Nov 26, 2017 at 17:29
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    $\begingroup$ @EricDuminil It means $f$ is the zero function, i.e. $f(x)=0$ for all $x$. $\endgroup$
    – wythagoras
    Commented Nov 26, 2017 at 19:44
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    $\begingroup$ Thanks. But why use a different notation for $f$ and $g$? $\endgroup$
    – Eric Duminil
    Commented Nov 26, 2017 at 19:49
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    $\begingroup$ @EricDuminil the expression for $g(x)$ involves $x$, so there is no ambiguity that $g(x) = \frac{1}{N} \sin N^2 x$ is a definition (as by convention of equality symbol). On the contrary, $f(x) = 0$ may be confused as an equation, rather than a definition, so sometimes people use $\equiv$ to emphasize they do mean $f(x)$ equals the constant function 0. $\endgroup$
    – Pig
    Commented Nov 26, 2017 at 22:02
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$f\sim g$ does not imply $f'\sim g'$! L'hopital's rule only works in one direction: $$\log x \sim \log \left((5+\sin x)x\right) \quad\text{but}\quad\frac1{x}\nsim\frac{((5+\sin x)x)'}{(5+\sin x)x}$$ or if you want, $$\log\log x \sim \log \log \left((5+\sin x)x\right) \quad\text{but}\quad\frac1{x\log x}\nsim\frac{((5+\sin x)x)'}{(5+\sin x)x \cdot \log\left((5+\sin x)x\right)}$$ (The factor $5+\sin x$ is there just to make the second quotient misbehave.)

The point is that we don't know (a priori) that $$\frac{\pi(x)}{x/\log x}$$ has a limit for $x\to\infty$.

What l'Hopital does tell us, is that if the limit of $(\pi(x)\log x)/x$ exists, then it is $1$.

I believe Chebyshev's original proof (and any subsequent ones) of this fact also goes along these lines, via a Mertens-type estimate for $\sum_{p\leq x}1/p\sim\int_1^x\pi(t)/t^2$.

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    $\begingroup$ Good addition that it does give a valid proof that if the limit from the Prime Number Theorem exists, then it is equal to $1$. The biography page http://www-history.mcs.st-andrews.ac.uk/Biographies/Chebyshev.html confirms that Chebyshev (Чебышёв) had proved this (around 1850 when his results were strong enough to prove Bertrand's postulate). That the limit actually exists was not proved until 1896, two years after his death. $\endgroup$
    – Jeppe Stig Nielsen
    Commented Nov 27, 2017 at 9:30

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