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In [1] Wikipedia say that for $\Re s>1$ the Riemann zeta function satisfies $$\log \zeta(s)=s\int_0^\infty\frac{\pi(x)}{x(x^s-1)}dx,$$ where $\pi(x)$ is the prime counting function, and say too that such expression is related with Prime Number Theorem (I don't know the history of this equation).

I've made the change $u=x^{1+\frac{1}{p_n}}$ (see below for details), where thus $s=1+\frac{1}{p_n}$ and $p_n\geq 2$ is the nth prime number (first, a fixed prime number, after I want test $n\to\infty$) and with this easy idea we can write, if there are no mistakes

$$\log\zeta(1+\frac{1}{p_n})=\int_0^\infty \frac{\pi(u^{\frac{p_n}{1+p_n}})}{u(u-1)}du.$$

My question, with the goal to learn and refresh my mathematics is about (see below for more details):

Question. Is right the change of variable in the step function $\pi(x)$ under the integral sign , this is, how do you formalize this step? What about $\lim_{n\to\infty}\int_0^\infty \frac{\pi(u^{p_n/(1+p_n)})}{u(u-1)}du?$ Since I believe that it is possible a formalized proof that $\lim_{n\to\infty} \log \zeta(1+1/p_n)=\infty$ then previous integral should be divergent. Thanks in advance.

My attempt was (firstly the philosophy is to try made an answer related with a nice topic, in the way to embroil topics, since Wikipedia say that previous identity is related with Prime Number theorem I've used primes) from the change $u=x^s=x^{1/(1+p_n)}$ compute $du=(1+1/p_n)x^{1/p_n}$, thus from $\frac{u}{x}=x^{1/p_n}$ we can write $$dx=\frac{1}{1+\frac{1}{p_n}}\frac{x}{u}du,$$ and then we compute the integral as $$\int_0^\infty \frac{\pi(u^{\frac{p_n}{1+p_n}})}{(1+\frac{1}{p_n})u(u-1)}du,$$ if there are no mistakes, and I claim that since the prime-countig function is a function whose set of discontinuities is enumerable (primes) its Lebesgue measure is zero (I don't know if it is wrong or yes is a formalized proof that the change of variable is feasible).

After we know Prime Number Theorem in the equivalents assertions (see [2]): $\pi(x)\sim\frac{x}{\log x}$ as $x\to\infty$, and $p_n\sim n\log n$ as $n\to\infty$. Too we know that $\zeta(1)=\sum_{n=1}^\infty \frac{1}{n}$ is divergent, and I claim that I can take the limit (I don't know if it is posible from proved identity involving $\log\zeta(1+\frac{1}{p_n})$, since at the infinity $\Re(1+\frac{1}{p_n})$ is equal to $1$), and claim that the integral diverges since $\log\infty=\infty$. When I've used Prime Number Theorem, we can write if there are no mistakes the asymptotic equivalences $$\pi(u^{\frac{p_n}{1+p_n}})\sim\frac{u^1}{u(u-1)(\frac{p_n}{1+p_n}\log u)}=\frac{1}{(\frac{p_n}{1+p_n})(u-1)\log u},$$ and claim, if it is right, previous statement since the integral $\int_0^\infty\frac{dt}{(t-1)\log t}$ does no converge. I want the formalized details of previous computations to obtain the best abilities and refresh my mathematics.

References (for example):

[1] Wikipedia, Riemann zeta function, section Mellin transform, https://en.wikipedia.org/wiki/Riemann_zeta_function

[2] Apostol, Introduction to Analytic Number Theory, Chapter 4.

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  • $\begingroup$ Please if there are mistakes tell me to delete these and improve my question without these. Too my thoughts were try to embroil with the equivalence with RH : $\forall \epsilon>0$ is $\pi(y)=\int_2^y dt/log t +O(y^{1/2+\epsilon})$ but it is out of my scope. $\endgroup$ – user243301 Nov 21 '15 at 15:52
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    $\begingroup$ The presence of $\ln\zeta(s)$ begs for Euler's infinite product representation of the $\zeta$ function. Split $(0,\infty)$ into intervals of the form $\big(p_k,~p_{k+1}\big)$. $\endgroup$ – Lucian Nov 21 '15 at 20:28
  • $\begingroup$ This was not known for me, thanks @Lucian $\endgroup$ – user243301 Nov 21 '15 at 20:30

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