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I think I found a way to write the truncated Prime $\zeta$ function like this: $$ P_x(s)=\sum_{p<x} \frac{1}{p^s} =\sum_{n=1}^{\infty}\frac{ \mu (n)}{n} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac zn-s}) \right]^{x}_2 \tag{7}, $$ see here for confirmation (or please correct me if I'm wrong)!

My next step of interest would be to know something about its Fourier Transform.

EDIT, containing some fresh ideas, changing the flavour of the question

The answer seems trivial when you look at the lhs of $(7)$, set $s\to is$ and rewrite it as $$ P_x(is)=\sum_{p<x} \frac{1}{p^{is}} = \sum_{p<x} \exp(-is\ln p) $$ to see that spectrum is (remember that $x$ is fixed) $$ \begin{eqnarray} \int_{-\infty}^\infty P_x(is)\exp(i2\pi s \omega) ds &=&\sum_{p<x} \int_{-\infty}^\infty \exp(-is\ln p) \exp(i2\pi s \omega)ds \\ &=&\sum_{p<x} \int_{-\infty}^\infty \exp\biggr(is(2\pi \omega-\ln p)\biggr) ds\\ &=&\sum_{p<x}\delta( \omega -\ln p)\tag{8} \end{eqnarray} $$

But when I take Fourier's integral deeper into the ${\bf r}\text{abbit's } {\bf h}\text{ole} {\bf s}$ of $(7)$, I'll get $$ \begin{eqnarray} \color{green}{\int_{-\infty}^\infty ds e^{i2\pi s \omega}}P_x(is) &=& \sum_{n=1}^{\infty}\frac{ \mu (n)}{n} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \underbrace{ \left( \color{green}{\int_{-\infty}^\infty ds e^{i2\pi s \omega}} \left[{\rm li}(t^{\frac zn-s}) \right]^{x}_2 \right) }_{\uparrow} \tag{9} \end{eqnarray} $$

But as robjohn pointed out in the comments to my (attempt of an) answer this integral is unlikely to converge. But we have $(8)$.

Does this mean that the sum of sums, $\sum_{n=1}^{\infty}\frac{ \mu (n)}{n} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}}$, forces convergence?


OLD ATTEMPT

I tried the following:

  1. Write it down for one of the summands ${\rm li}(x^{y})$ with $y=\frac zn-s\in \Bbb C$: $$ \int_{-\infty}^\infty \exp(2\pi i\omega y){\rm li}(x^{y}) dy \tag{*} $$
  2. Use partial integration $\int u' v = [uv]-\int uv'$ to get $$ \begin{eqnarray} \int_{-\infty}^\infty \exp(2\pi i\omega y){\rm li}(x^{y}) dy&=& \left[\frac{\exp(2\pi i\omega y)}{2\pi i\omega}{\rm li}(x^{y})\right]_{-\infty}^\infty-\int_{-\infty}^\infty \frac{\exp(2\pi i\omega y)x^y\log x}{\log(x^{y})} dy\\ \end{eqnarray} $$
  3. I'm not sure about the $[\cdots]_{-\infty}^\infty$-parts. I hope it vanishes. Is this true?
  4. My next idea was to Taylor expand $\exp(2\pi i\omega y)=\sum_{k=0}^\infty \frac{(2\pi i \omega)^k}{k!}y^k$ to get: $$ -\sum_{k=0}^\infty \frac{(2\pi i \omega)^k}{k!} \int_{-\infty}^\infty \frac{y^k x^y\log x}{\log(x^{y})} dy $$ Our good friend Wolfram can help with the last integral ($x$ and $y$ interchanged for the sake of simpler inputs for W|A!!!) in an indefinite form for fixed $k=$ $1$, $2$ or $3$, but I think I'm kind of led astray.

Even with an analytical expression for that I'm not sure if this is the way to go. Would it be better to expand ${\rm li}(x^y)$?

Can anyone help me?

Thanks,

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1 Answer 1

This was my first attempt to answer the old question. It might be wrong, but I'll keep it due to the valuable comments below:

Use ${\rm li}(z)={\rm Ei}(\ln z)$ to get:

$$ \begin{eqnarray} \int_{-\infty}^\infty \exp(2\pi i\omega y){\rm li}(x^{y}) dy&=& \int_{-\infty}^\infty \exp(2\pi i\omega y){\rm Ei}(y\ln x) dy\\ &=&\left[\frac{\exp(2\pi i\omega y)}{2\pi i\omega}{\rm Ei}(y\ln x)\right]_{-\infty}^{\phantom{-}\infty} -\int_{-\infty}^\infty \exp(2\pi i\omega y)\frac{\exp(y\ln x)}{y} dy\\ &=&\left[\frac{\exp(2\pi i\omega y)}{2\pi i\omega}{\rm Ei}(y\ln x)\right]_{-\infty}^{\phantom{-}\infty} -\frac{{\rm Ei}\biggr(y(2\pi i \omega-\ln x)\biggr)}{2\pi i\omega}, \end{eqnarray} $$

where Wolfram helped a bit. Assuming a vanishing bracket, this gives a nice result...

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How can $\int_{-\infty}^\infty \exp(2\pi i\omega y)\frac{\exp(y\ln x)}{y} dy$ converge? Whether $\ln(x)$ is positive or negative, one side of the integral should diverge. –  robjohn Jan 21 '13 at 11:44
    
@robjohn so you say that the Fourier Transform doesn't exist? How else can I get some information about the periodic properties of $li(\cdot)$? –  draks ... Jan 21 '13 at 11:46
    
That doesn't follow, unless you know that the bracket quantity is finite, which it isn't. $\mathrm{Ei}(x)$ blows up like $e^x/x$ as $x\to+\infty$. That is, the final answer above is $\infty-\infty$. –  robjohn Jan 21 '13 at 11:52
    
Isn't the bracket $\infty-0$ since $\lim_{y\to-\infty} Ei(y)=0$? What if I restricted my region of interest to let's say $y<y_\max$ by using some kind of boxing (is this the right word...?) function? –  draks ... Jan 21 '13 at 11:56
    
Yes, the bracket is $\infty$ and the integral is $\infty$. Adding some cutoff function can strongly affect the Fourier Transform, but it might work. –  robjohn Jan 21 '13 at 11:58
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