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Is this Riemann integral for prime numbers true?
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While answering another question in MSE, I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof.
Let $p_k$ be the $k$-th prime and let $f$ be a continuous function Riemann
integrable in $(0,1)$ then,
$$ lim_n to inftyfrac1nsum_r = 1^nfBig(fracp_rp_nBig) = int_0^1f(x)dx. $$
My proof was based on showing that as $n to infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a trivial property of equidistributed sequence.
Motivation: I have found countless identities, limits etc on prime numbers using this simple formula, including all answers to all three questions on the arithmetic, geometric and harmonic means of primes mentioned in the above link.
integration number-theory analysis definite-integrals prime-numbers
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
 |Â
show 4 more comments
up vote
7
down vote
favorite
While answering another question in MSE, I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof.
Let $p_k$ be the $k$-th prime and let $f$ be a continuous function Riemann
integrable in $(0,1)$ then,
$$ lim_n to inftyfrac1nsum_r = 1^nfBig(fracp_rp_nBig) = int_0^1f(x)dx. $$
My proof was based on showing that as $n to infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a trivial property of equidistributed sequence.
Motivation: I have found countless identities, limits etc on prime numbers using this simple formula, including all answers to all three questions on the arithmetic, geometric and harmonic means of primes mentioned in the above link.
integration number-theory analysis definite-integrals prime-numbers
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
2
If you take the example $f(x)=2x$, then the right-hand side is just one. However, for a huge (fixed) value of $n$, there would be more primes in the interval $[0,p_n/2]$ than in the interval $[p_n/2,p_n]$, so why whould the value of that average be near the number one? ADDITION: As an example, among the first half billion numbers there are 26355867 primes, but among the next half billion numbers, i.e. between $0.5cdot 10^9$ and $10^9$, there are only 24491667 primes.
– Jeppe Stig Nielsen
2 hours ago
1
@JeppeStigNielsen How do you know that the first interval will have more primes asymptotically?
– Yanko
2 hours ago
1
@JeppeStigNielsen I think the same as you. For me, it would look nicer, if one changes $dx$ by another integrator (kinda Riemann-Stieltjes) related to the distribution of primes.
– Marcuswood
2 hours ago
2
@JeppeStigNielsen use the prime number theorem: $pi(n)/pi(2n)approx (n/log (n))/(2nlog(2n)) = 1/2 (log(2)+log(n))/log(n)to 1/2$
– Bananach
2 hours ago
1
@Marcuswood the prime number theorem says that F(x)=x indeed. This is essentially because the "average gap" $log x$ between prime numbers is a slowly varying function, too slow to matter when the numbers of prime numbers at $n$ and $x n$ are considered with $ntoinfty$.
– Bananach
2 hours ago
 |Â
show 4 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
While answering another question in MSE, I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof.
Let $p_k$ be the $k$-th prime and let $f$ be a continuous function Riemann
integrable in $(0,1)$ then,
$$ lim_n to inftyfrac1nsum_r = 1^nfBig(fracp_rp_nBig) = int_0^1f(x)dx. $$
My proof was based on showing that as $n to infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a trivial property of equidistributed sequence.
Motivation: I have found countless identities, limits etc on prime numbers using this simple formula, including all answers to all three questions on the arithmetic, geometric and harmonic means of primes mentioned in the above link.
integration number-theory analysis definite-integrals prime-numbers
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
While answering another question in MSE, I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof.
Let $p_k$ be the $k$-th prime and let $f$ be a continuous function Riemann
integrable in $(0,1)$ then,
$$ lim_n to inftyfrac1nsum_r = 1^nfBig(fracp_rp_nBig) = int_0^1f(x)dx. $$
My proof was based on showing that as $n to infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a trivial property of equidistributed sequence.
Motivation: I have found countless identities, limits etc on prime numbers using this simple formula, including all answers to all three questions on the arithmetic, geometric and harmonic means of primes mentioned in the above link.
integration number-theory analysis definite-integrals prime-numbers
integration number-theory analysis definite-integrals prime-numbers
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
edited 1 hour ago
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
asked 3 hours ago
Nilotpal Kanti Sinha
3,61021334
3,61021334
2
If you take the example $f(x)=2x$, then the right-hand side is just one. However, for a huge (fixed) value of $n$, there would be more primes in the interval $[0,p_n/2]$ than in the interval $[p_n/2,p_n]$, so why whould the value of that average be near the number one? ADDITION: As an example, among the first half billion numbers there are 26355867 primes, but among the next half billion numbers, i.e. between $0.5cdot 10^9$ and $10^9$, there are only 24491667 primes.
– Jeppe Stig Nielsen
2 hours ago
1
@JeppeStigNielsen How do you know that the first interval will have more primes asymptotically?
– Yanko
2 hours ago
1
@JeppeStigNielsen I think the same as you. For me, it would look nicer, if one changes $dx$ by another integrator (kinda Riemann-Stieltjes) related to the distribution of primes.
– Marcuswood
2 hours ago
2
@JeppeStigNielsen use the prime number theorem: $pi(n)/pi(2n)approx (n/log (n))/(2nlog(2n)) = 1/2 (log(2)+log(n))/log(n)to 1/2$
– Bananach
2 hours ago
1
@Marcuswood the prime number theorem says that F(x)=x indeed. This is essentially because the "average gap" $log x$ between prime numbers is a slowly varying function, too slow to matter when the numbers of prime numbers at $n$ and $x n$ are considered with $ntoinfty$.
– Bananach
2 hours ago
 |Â
show 4 more comments
2
If you take the example $f(x)=2x$, then the right-hand side is just one. However, for a huge (fixed) value of $n$, there would be more primes in the interval $[0,p_n/2]$ than in the interval $[p_n/2,p_n]$, so why whould the value of that average be near the number one? ADDITION: As an example, among the first half billion numbers there are 26355867 primes, but among the next half billion numbers, i.e. between $0.5cdot 10^9$ and $10^9$, there are only 24491667 primes.
– Jeppe Stig Nielsen
2 hours ago
1
@JeppeStigNielsen How do you know that the first interval will have more primes asymptotically?
– Yanko
2 hours ago
1
@JeppeStigNielsen I think the same as you. For me, it would look nicer, if one changes $dx$ by another integrator (kinda Riemann-Stieltjes) related to the distribution of primes.
– Marcuswood
2 hours ago
2
@JeppeStigNielsen use the prime number theorem: $pi(n)/pi(2n)approx (n/log (n))/(2nlog(2n)) = 1/2 (log(2)+log(n))/log(n)to 1/2$
– Bananach
2 hours ago
1
@Marcuswood the prime number theorem says that F(x)=x indeed. This is essentially because the "average gap" $log x$ between prime numbers is a slowly varying function, too slow to matter when the numbers of prime numbers at $n$ and $x n$ are considered with $ntoinfty$.
– Bananach
2 hours ago
2
2
If you take the example $f(x)=2x$, then the right-hand side is just one. However, for a huge (fixed) value of $n$, there would be more primes in the interval $[0,p_n/2]$ than in the interval $[p_n/2,p_n]$, so why whould the value of that average be near the number one? ADDITION: As an example, among the first half billion numbers there are 26355867 primes, but among the next half billion numbers, i.e. between $0.5cdot 10^9$ and $10^9$, there are only 24491667 primes.
– Jeppe Stig Nielsen
2 hours ago
If you take the example $f(x)=2x$, then the right-hand side is just one. However, for a huge (fixed) value of $n$, there would be more primes in the interval $[0,p_n/2]$ than in the interval $[p_n/2,p_n]$, so why whould the value of that average be near the number one? ADDITION: As an example, among the first half billion numbers there are 26355867 primes, but among the next half billion numbers, i.e. between $0.5cdot 10^9$ and $10^9$, there are only 24491667 primes.
– Jeppe Stig Nielsen
2 hours ago
1
1
@JeppeStigNielsen How do you know that the first interval will have more primes asymptotically?
– Yanko
2 hours ago
@JeppeStigNielsen How do you know that the first interval will have more primes asymptotically?
– Yanko
2 hours ago
1
1
@JeppeStigNielsen I think the same as you. For me, it would look nicer, if one changes $dx$ by another integrator (kinda Riemann-Stieltjes) related to the distribution of primes.
– Marcuswood
2 hours ago
@JeppeStigNielsen I think the same as you. For me, it would look nicer, if one changes $dx$ by another integrator (kinda Riemann-Stieltjes) related to the distribution of primes.
– Marcuswood
2 hours ago
2
2
@JeppeStigNielsen use the prime number theorem: $pi(n)/pi(2n)approx (n/log (n))/(2nlog(2n)) = 1/2 (log(2)+log(n))/log(n)to 1/2$
– Bananach
2 hours ago
@JeppeStigNielsen use the prime number theorem: $pi(n)/pi(2n)approx (n/log (n))/(2nlog(2n)) = 1/2 (log(2)+log(n))/log(n)to 1/2$
– Bananach
2 hours ago
1
1
@Marcuswood the prime number theorem says that F(x)=x indeed. This is essentially because the "average gap" $log x$ between prime numbers is a slowly varying function, too slow to matter when the numbers of prime numbers at $n$ and $x n$ are considered with $ntoinfty$.
– Bananach
2 hours ago
@Marcuswood the prime number theorem says that F(x)=x indeed. This is essentially because the "average gap" $log x$ between prime numbers is a slowly varying function, too slow to matter when the numbers of prime numbers at $n$ and $x n$ are considered with $ntoinfty$.
– Bananach
2 hours ago
 |Â
show 4 more comments
1 Answer
1
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up vote
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Too long for a comment
Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],cdots,[p_n-1/p_n,1]$$
Then, using Riemann sum, we have
$$I:=int^1_0f(x)dx=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)fracp_k+1-p_kp_n$$
If we assume that $p_j=jln j$,
$$I=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)h(k,n)+lim_ntoinftyfrac1n sum^n_k=1fleft(fracp_kp_nright) qquad(1)$$
where
$$h(k,n)=frac(k+1)ln(k+1)-kln knln n-frac1n$$
It can be shown that $$h(k,n)le h(n,n)=O(frac1nln n)$$
Therefore, the absolute value of the first term in $(1)$ is upper bounded by
$$h(n,n)cdot nMto 0$$ where $M$ is a positive constant. This leads us to our desired result.
I am not sure if this argument can be made rigorous. I will review it when I have leisure time.
answered 2 hours ago
Szeto
4,9811623
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Too long for a comment
Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],cdots,[p_n-1/p_n,1]$$
Then, using Riemann sum, we have
$$I:=int^1_0f(x)dx=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)fracp_k+1-p_kp_n$$
If we assume that $p_j=jln j$,
$$I=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)h(k,n)+lim_ntoinftyfrac1n sum^n_k=1fleft(fracp_kp_nright) qquad(1)$$
where
$$h(k,n)=frac(k+1)ln(k+1)-kln knln n-frac1n$$
It can be shown that $$h(k,n)le h(n,n)=O(frac1nln n)$$
Therefore, the absolute value of the first term in $(1)$ is upper bounded by
$$h(n,n)cdot nMto 0$$ where $M$ is a positive constant. This leads us to our desired result.
I am not sure if this argument can be made rigorous. I will review it when I have leisure time.
answered 2 hours ago
Szeto
4,9811623
add a comment |Â
up vote
6
down vote
Too long for a comment
Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],cdots,[p_n-1/p_n,1]$$
Then, using Riemann sum, we have
$$I:=int^1_0f(x)dx=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)fracp_k+1-p_kp_n$$
If we assume that $p_j=jln j$,
$$I=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)h(k,n)+lim_ntoinftyfrac1n sum^n_k=1fleft(fracp_kp_nright) qquad(1)$$
where
$$h(k,n)=frac(k+1)ln(k+1)-kln knln n-frac1n$$
It can be shown that $$h(k,n)le h(n,n)=O(frac1nln n)$$
Therefore, the absolute value of the first term in $(1)$ is upper bounded by
$$h(n,n)cdot nMto 0$$ where $M$ is a positive constant. This leads us to our desired result.
I am not sure if this argument can be made rigorous. I will review it when I have leisure time.
answered 2 hours ago
Szeto
4,9811623
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Too long for a comment
Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],cdots,[p_n-1/p_n,1]$$
Then, using Riemann sum, we have
$$I:=int^1_0f(x)dx=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)fracp_k+1-p_kp_n$$
If we assume that $p_j=jln j$,
$$I=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)h(k,n)+lim_ntoinftyfrac1n sum^n_k=1fleft(fracp_kp_nright) qquad(1)$$
where
$$h(k,n)=frac(k+1)ln(k+1)-kln knln n-frac1n$$
It can be shown that $$h(k,n)le h(n,n)=O(frac1nln n)$$
Therefore, the absolute value of the first term in $(1)$ is upper bounded by
$$h(n,n)cdot nMto 0$$ where $M$ is a positive constant. This leads us to our desired result.
I am not sure if this argument can be made rigorous. I will review it when I have leisure time.
answered 2 hours ago
Szeto
4,9811623
Too long for a comment
Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],cdots,[p_n-1/p_n,1]$$
Then, using Riemann sum, we have
$$I:=int^1_0f(x)dx=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)fracp_k+1-p_kp_n$$
If we assume that $p_j=jln j$,
$$I=lim_ntoinftysum^n_k=1fleft(fracp_kp_nright)h(k,n)+lim_ntoinftyfrac1n sum^n_k=1fleft(fracp_kp_nright) qquad(1)$$
where
$$h(k,n)=frac(k+1)ln(k+1)-kln knln n-frac1n$$
It can be shown that $$h(k,n)le h(n,n)=O(frac1nln n)$$
Therefore, the absolute value of the first term in $(1)$ is upper bounded by
$$h(n,n)cdot nMto 0$$ where $M$ is a positive constant. This leads us to our desired result.
I am not sure if this argument can be made rigorous. I will review it when I have leisure time.
answered 2 hours ago
Szeto
4,9811623
edited 2 hours ago
answered 2 hours ago
Szeto
4,9811623
answered 2 hours ago
Szeto
4,9811623
answered 2 hours ago
Szeto
4,9811623
4,9811623
add a comment |Â
add a comment |Â
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2
If you take the example $f(x)=2x$, then the right-hand side is just one. However, for a huge (fixed) value of $n$, there would be more primes in the interval $[0,p_n/2]$ than in the interval $[p_n/2,p_n]$, so why whould the value of that average be near the number one? ADDITION: As an example, among the first half billion numbers there are 26355867 primes, but among the next half billion numbers, i.e. between $0.5cdot 10^9$ and $10^9$, there are only 24491667 primes.
– Jeppe Stig Nielsen
2 hours ago
1
@JeppeStigNielsen How do you know that the first interval will have more primes asymptotically?
– Yanko
2 hours ago
1
@JeppeStigNielsen I think the same as you. For me, it would look nicer, if one changes $dx$ by another integrator (kinda Riemann-Stieltjes) related to the distribution of primes.
– Marcuswood
2 hours ago
2
@JeppeStigNielsen use the prime number theorem: $pi(n)/pi(2n)approx (n/log (n))/(2nlog(2n)) = 1/2 (log(2)+log(n))/log(n)to 1/2$
– Bananach
2 hours ago
1
@Marcuswood the prime number theorem says that F(x)=x indeed. This is essentially because the "average gap" $log x$ between prime numbers is a slowly varying function, too slow to matter when the numbers of prime numbers at $n$ and $x n$ are considered with $ntoinfty$.
– Bananach
2 hours ago