# Integral involving Pi(x)

1. Aug 28, 2006

### lokofer

Hello..that's my question today..why can't we obtain $$\pi(x)$$ by solving the integral equation obtained from Euler's product:

$$\frac{log \zeta(s)}{s}= \int_{2}^{\infty}dx \frac{ \pi(x)}{x(x^{s}-1)}$$ ?

- Of course we can't solve it "Analytically" (or perhaps yes, i will take a look to "Numerical Recipes"...:grumpy: ) but we could solve it Numerically using some quadrature method for the Integral equation..or introducing the term inside the Kernel:

$$\pi(s) = \int_{2}^{\infty} \pi(x) \delta (x-s)$$ so the integral becomes a "Fredholm Integral Equation of Second Kind".... I know that an algorithm (either numerical or similar) must exist to solve any Integral equation Numerically...why not for the Prime counting function?...

2. Aug 28, 2006

### matt grime

Look, Jose, a naive algorithm exists for calculating pi(x), and one exists for evaluating integrals. Now, try to estimate for your self which is more expensive.

In the several years you've been posting here you have never managed to distinguish between a function and an algorithm for evaluating a function at a given point. Numerically you can do anything you want to anything you wish, but it all costs time and 'mathematical money'. So why don't you work out the cost of estimating firstly the functions you wish to use,and thence the integral, and then see if it is better or worse than just working out pi(x) by hand?

Here's a rough heuristic as to why what you've done doesn't seem like a saving:

pi(x) is a perfectly well defined 'analytic' expression. We can work out its values relatively cheaply.

you have replaced pi(x) with another function that is defined in terms of an infinite series, then you've used an integral that can only be done numerically, all of which 'cost money' to approximate, even if there were a way to recover pi(x) from that expression.

Last edited: Aug 28, 2006
3. Aug 28, 2006

### lokofer

You could use the Resolvent Kernel approximation to obtain [\tex] \pi(x) [/tex] :

$$\int_{2}^{\infty}ds R(x,s) \frac{log \zeta (s)}{s}= \pi(x)$$

Thta's what i would use..in fact Matt i'm not mathematician, i'm just a physicist suffering unemployment..i don't know much about algorithms but i think that with a "good" computer today you could calculate the "Resolvent" kernel by iterations or other method.. perhaps in a pair of months you could obtain R(x,s) the rest is easy..

4. Aug 28, 2006

### matt grime

And why can't a good computer just work out pi(x) directly even more quickly?

And you can't pull the 'I'm not a mathematician' line again. You've been given years of advice here, it is not our fault you choose to ignore it. It has been explained to you many many times, that just because you can rearrange some symbols does not make it easier to evaluate. It is up to you to work out the estimates of cost, and as a physicist your are in a far better position to do that than a mathematician (estimates, asmyptotics, numerics etc).

Last edited: Aug 28, 2006