W have $$\int_{2}^{x} dt d\pi (t) t^{2} = \sum_{p \le x}p^{2}$$

also for every prime p then $$\sigma _{2} (p) = 1+p^{2}$$

by the definition of 'divisor function' of order 2

so $$\sum_{p \le x}p^{2}+ \pi (x) = \int_{2}^{x} dt d\pi (t) \sigma_{2}(t) = \sum_{p \le x} \sigma _{2} (p)$$

since for every prime the divisor function has only 2 numbers 1 and p then differentiating to both sides we find:

$$d \pi (x) x^{2} = d \pi (x) \sigma_{2} (x)+ d \pi(x)$$

which is completely absurd since we could remove the derivative of the prime counting function..i believe that perhaps a derivative of second order $$d^{2} \pi (x)$$ or a factor $$d \pi (x) d \pi (x)$$ should appear.

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Hurkyl
Staff Emeritus
Gold Member
$$\int_{2}^{x} dt d\pi (t) t^{2}$$
This expression makes no sense. What did you mean to write?

matt grime
Homework Helper
Who knows. It is Jose (eljose), after all. I mean, who thinks that the 'prime counting function' is differentiable, apart from him?

He keeps making new accounts.

Since the 'Prime counting function' is an step function.then its derivative can be considered in the sense of 'distribution theory. and

$$\int_{2}^{x} dt d\pi (t) f(t)= \sum _{p \le x} f(p)$$

integrating by parts you get the usual expression due to Abel sum formula.

$$\frac{d \pi (x) }{dx}= \sum_{p} \delta (x-p)$$

here the sum is extended over all primes ..then could somebody respond to my 'paradox' ??

matt grime
Homework Helper
Let's start by calling it a 'mistake', not a paradox. Now, if you were to write it out clearly, using sentences, perhaps someone will attempt to work out what you're up to. One obvious thing is you want to 'remove the derivative of ...'. If by that you mean divide by a distribution that is almost always zero, you're bound to introduce oddities.

But..how the hell can you multiply or divide distributions?'..in fact:

$$\delta (x) \delta (x) = \frac{Sin(Nx)^{2}}{x^{2}}$$ (1)

hence, Schwartz theorem is false.. the same for any other distribution..let be a sucession of function then

$$f_{n} (x)g_{n} (x) \rightarrow S(x)U(x)$$ (2)

in both cases (1) and (2) you must take n-->oo as you can see the theorem by Schwartz is false.

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matt grime
Homework Helper
You're the person who wrote 'remove the...' without saying what you meant. Once more eljose, you're just ranting away and not making yourself at all intelligible.

sorry in this case it was my fault by 'remove' i meant divide each term by $$d \pi (x)$$ instead i don't understand why this 'mess' about distributions..whenever you consider them as a limit of a certain sucession of function it's all clearer...don't you think also reading 'Calculus' by Spivak the notation is not new to prove Euler's sum formula they take:

$$\int _{0}^{\infty}f(x)d([x]) = \sum_{n=0}^{\infty} f(n)$$

matt grime