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I was thinking about the following:

Denote $\pi(x)$ as the prime counting function such that: $$ \pi(x) = \#\text{ of prime numbers}\leq x $$ It is well known from the prime number theorem that $$ \pi(x) \sim \frac{x}{\ln x} $$ and $$ \pi(x) \sim \text{Li}(x),\quad \text{Li}(x)=\int_2^x\frac{1}{\ln t}\,dt. $$ Note the following: if $A(x) \sim B(x)\implies A(x)/B(x) = 1 \text{ as }x\to\infty$ and $A(x), B(x)\to\infty \text{ as }x\to\infty$.

If $C(x)$ is a function such that $n \leq C(x) \leq k$ then:

$$(A(x) + n)/B(x) \leq (A(x) + C(x))/B(x) \leq (A(x) + k)/B(x)$$

$$A(x)/B(x) + n/B(x) \leq A(x)/B(x) + C(x)/B(x) \leq A(x)/B(x) + k/B(x)$$

If x is taken to infinity:

$$1 + 0 \leq A(x)/B(x) + C(x)/B(x) \leq 1 + 0$$

$$\rightarrow A(x) + C(x) ~ B(x)$$

What interests me is that since we already know that:

$$\pi(x) \sim x/\ln(x)$$

and from above that $x/ln(x) +$ any number of functions of the form $A(b(x)) ~ \pi(x)$

Can we not try to do some sort of fourier analysis on the function:

$$\pi(x) - x/\ln(x) $$

or

$\pi(x) - \text{Li}(x)$?

NasuSama
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Sidharth Ghoshal
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    I fail to understand what your saying at the point in which you introduce the function $c(x)$, I recommend formatting your text properly. – Ethan Splaver May 05 '13 at 23:33
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    I don't understand what you mean by " do some fourier analysis on the function $\pi(x)-\frac{x}{\ln(x)}$ ". If you are asking for a good estimate on the difference between $\pi(x)$ and $\frac{x}{\ln(x)}$, then this would be asymptotic to $\text{Li}(x)-\frac{x}{\ln(x)}$, this is because the logarithmic integral is a much better approximation to $\pi(x)$ then $\frac{x}{\ln(x)}$, so with that said it would probably be more interesting to study the behavior of $\pi(x)-\text{Li(x)}$. – Ethan Splaver May 05 '13 at 23:46
  • basically, I want to see if there are any patterns or noticeable trends in the difference, that can be approximated by sines – Sidharth Ghoshal May 05 '13 at 23:51
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    The difference between $\pi(x)$ and $\frac{x}{\ln(x)}$ can already be approximated very well, I think the question to ask would be regarding the difference of $\pi(x)$ and $\text{Li}(x)$. – Ethan Splaver May 05 '13 at 23:52
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    The explicit formula for the prime counting function is perhaps the 'correct' mixture of fourier analysis (using the zeros of the Riemann zeta function) and asymptotics (employing the logarithmic integral function). The PNT $\pi(x)\sim x/\log x$ is equivalent to $\psi(x)\sim x$, where $\psi(x)$ is the chebyshev function, a variant way of measuring prime growth. A more direct fourier expansion of $\psi(x)-x$ also exists, a variant of the explicit formula. See the link for more details. – anon May 06 '13 at 01:20

2 Answers2

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The purpose of $\zeta(s)$ is to perform Fourier analysis on $\pi(e^u)$ :

$$\zeta(s) = \prod_p \frac{1}{1-p^{-s}} \quad \implies \quad \ln \zeta(s) = \sum_{p^k} \frac{p^{-sk}}{k}$$

i.e. it is the $\color{red}{\text{Laplace transform}}$ of $\sum_{p^k} \frac{1}{k}\delta(u-\ln p^k)$ $\qquad \scriptstyle \text{(or if you prefer, }\hat{f}(\xi ) =\ln \zeta(\sigma+2i \pi \xi) \text{ is the Fourier transform of } f(u) = \sum_{p^k} \frac{1}{\scriptstyle k \ p^{ \sigma k}}\delta(u-\ln p^k)$)

Integrating by parts, you get $$\ln \zeta(s) = s \int_0^\infty J(e^u) e^{-su}du$$ where $J(x) = \sum_{p^k < x}\frac{1}{k} = \pi(x) + \mathcal{O}(\sqrt{x})$.

i.e. $$\boxed{\frac{\ln \zeta(s) }{s} = h(s) + \int_0^\infty \pi(e^u) e^{-su}du}$$ where $h(s) = \int_0^\infty (J(e^u)-\pi(e^u))e^{-su}du$ is analytic and bounded for $Re(s) > 1/2+\epsilon$.

reuns
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  • I'm sorry, please can you tell me if your integral expression for $\log \zeta(s)$ is the same integral identity $$\ln \zeta(s) = s \int_0^\infty J(x) x^{-s-1}dx,$$ from Wikipedia's article for the Riemann Zeta? –  Sep 07 '16 at 16:35
  • @user243301 yes of course $x = e^u$ – reuns Sep 07 '16 at 16:37
  • Sorry, since were mistakes in my calculations with such change of variable. Now thanks your hint I see those factors in the integrand, but how do you solve the lower limit of integration $\int_0 du$ : I say when one has $x=0$, using such change of variable I need to solve the $0=e^u$. How do you know that your lower limit of integration is $0=u$? –  Sep 07 '16 at 16:48
  • @user243301 I meant $\int_{\ln 2}^\infty J(e^u) e^{-su} du$, since $J(x) = 0$ for $x < 2$ – reuns Sep 07 '16 at 16:50
  • Sorry for my insistence, and thanks for your explanation. –  Sep 07 '16 at 16:52
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You can perform Fourier analysis on $\pi(x)$. Here are two plots of a Fourier series for $\pi'(x)$, i.e. the derivative of $\pi(x)$, with a frequency evaluation limit $f=1$ for the first plot and $f=4$ for the second plot. The two plots below illustrate the general relationship that this Fourier series always evaluates to $2f$ times the step size of $\pi(x)$ at integer values of $x$.

Fourier Series for π'(x) Evaluated at f=1

Fourier Series for π'(x) Evaluated at f=4

Björn Friedrich
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Steven Clark
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  • the Fourier series of $\pi'(x)$ ? I don't follow you. You meant the discrete Fourier transform of $x(p) = 1, x(n) = 0$ otherwise, $n \in 1 \ldots N$ ? – reuns Sep 07 '16 at 13:09