Sum over primes asymtotics

[zetafunction]

is the following asymptotic approximation valid whenever dealing sums over primes ?? $$\sum_{p\le x}f(p) \sim \int_{2}^{x}\frac{f(x)dx}{log(x)}$$

 

[Office_Shredder]

f(x)=1 if x is a prime number, 0 otherwise

You'll definitely need at least some assumptions like continuity, and probably uniform continuity, otherwise you could just make f(x) approximately zero except for at prime numbers, where it forms sharper and sharper spikes over time

Even then, if f(x) only had finite support, say f(x)=1 for x<6, f(x)=7-x if 6<x<7, 0 if x>7, then you will never get the difference to become small or the ratio of the two to become close to each other (I'm not sure exactly what you mean by ~ in this case but since for x>7 the two stop changing and are not equal to each other, it's obviously not the case that they become asymptotically close for any definition of asymptotic)

 

[zetafunction]

i mean ASYMPTOTIC that is the quotient between the series and the integral will tend to 1 for big x

for example if you set f(x)=1 you get the Prime Number Theorem

 

[Office_Shredder]

And if you set f(x) to be either of the examples that I gave in my post you don't get such a result

 

[CRGreathouse]

I assume the integral was intended to be $$\int_2^x\frac{f(t)dt}{\log t}$$.Here's a uniformly continuous function supported on x > 0 where the two differ asymptotically:

$$f(x) = 2^{-x}$$ for x a positive integer
$$f(x) = 1$$ for x a positive half-integer
$$f(x) = (\lfloor x\rfloor-x+1/2)f(\lfloor x\rfloor)+x-\lfloor x\rfloor$$ for $$\lfloor x\rfloor<x<\lfloor x\rfloor+1/2$$
$$f(x) = x-\lfloor x\rfloor-1/2+(\lfloor x\rfloor+1-x)f(\lfloor x\rfloor+1)$$ for $$\lfloor x\rfloor+1/2<x<\lfloor x\rfloor+1$$

$$\sum_{p\le x}f(p)\sim1\ldots$$ (actually, 0.4146825... + o(1), where the number is Sloane's A051006)
while
$$\int_2^x\frac{f(t)dt}{\log t}\sim\frac{x}{2\log x}$$What about restricting f to monotonic functions? I mean with all the other stuff -- without uniform continuity I think I can make it fail even for monotonics.

 

[zetafunction]

well, how about trying it for WELL BEHAVED function ? like polynomials , trigonometric and all this stuff ..