upper bound for integration of log of primes
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I want to give an upper bound for
$$int limits_{a}^{b} log{p_t} dt,$$
where $p_t$ is the $t$-th prime number, and one can generalize the idea of $p_t$ even for $t$ real numbers, simply by $p_t = p_{lfloor t rfloor}$ or any other approximation.
I think I have a good approximation, but I could not prove its an upper bound,
$$int limits_{a}^{b} ln{p_t} dt = F(b)-F(a),$$
where $F(x) = -x + x ln{(x ln {x} +2x ln ln x)}-li(x)$.
Could any one prove this is an upper bound, or give an upper bound with proof it's an upper bound.
integration prime-numbers approximation upper-lower-bounds
edited Nov 21 at 12:51
Glorfindel
3,38471730
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
add a comment |
up vote
0
down vote
favorite
I want to give an upper bound for
$$int limits_{a}^{b} log{p_t} dt,$$
where $p_t$ is the $t$-th prime number, and one can generalize the idea of $p_t$ even for $t$ real numbers, simply by $p_t = p_{lfloor t rfloor}$ or any other approximation.
I think I have a good approximation, but I could not prove its an upper bound,
$$int limits_{a}^{b} ln{p_t} dt = F(b)-F(a),$$
where $F(x) = -x + x ln{(x ln {x} +2x ln ln x)}-li(x)$.
Could any one prove this is an upper bound, or give an upper bound with proof it's an upper bound.
integration prime-numbers approximation upper-lower-bounds
edited Nov 21 at 12:51
Glorfindel
3,38471730
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
1
The integral expression is needlessly complicated. You are basically summing the log of successive primes. The prime number theorem (which is true) is equivalent to $sum_{p le x} log p sim x$ meaning the quotient of the LHS and the RHS tends to $1$ as $x to infty$.
– mathworker21
Dec 28 '17 at 6:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to give an upper bound for
$$int limits_{a}^{b} log{p_t} dt,$$
where $p_t$ is the $t$-th prime number, and one can generalize the idea of $p_t$ even for $t$ real numbers, simply by $p_t = p_{lfloor t rfloor}$ or any other approximation.
I think I have a good approximation, but I could not prove its an upper bound,
$$int limits_{a}^{b} ln{p_t} dt = F(b)-F(a),$$
where $F(x) = -x + x ln{(x ln {x} +2x ln ln x)}-li(x)$.
Could any one prove this is an upper bound, or give an upper bound with proof it's an upper bound.
integration prime-numbers approximation upper-lower-bounds
edited Nov 21 at 12:51
Glorfindel
3,38471730
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
I want to give an upper bound for
$$int limits_{a}^{b} log{p_t} dt,$$
where $p_t$ is the $t$-th prime number, and one can generalize the idea of $p_t$ even for $t$ real numbers, simply by $p_t = p_{lfloor t rfloor}$ or any other approximation.
I think I have a good approximation, but I could not prove its an upper bound,
$$int limits_{a}^{b} ln{p_t} dt = F(b)-F(a),$$
where $F(x) = -x + x ln{(x ln {x} +2x ln ln x)}-li(x)$.
Could any one prove this is an upper bound, or give an upper bound with proof it's an upper bound.
integration prime-numbers approximation upper-lower-bounds
integration prime-numbers approximation upper-lower-bounds
edited Nov 21 at 12:51
Glorfindel
3,38471730
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
edited Nov 21 at 12:51
Glorfindel
3,38471730
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
edited Nov 21 at 12:51
Glorfindel
3,38471730
edited Nov 21 at 12:51
Glorfindel
3,38471730
edited Nov 21 at 12:51
Glorfindel
3,38471730
3,38471730
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
asked Dec 28 '17 at 6:03
Ahmad
2,4921625
2,4921625
1
The integral expression is needlessly complicated. You are basically summing the log of successive primes. The prime number theorem (which is true) is equivalent to $sum_{p le x} log p sim x$ meaning the quotient of the LHS and the RHS tends to $1$ as $x to infty$.
– mathworker21
Dec 28 '17 at 6:07
add a comment |
1
The integral expression is needlessly complicated. You are basically summing the log of successive primes. The prime number theorem (which is true) is equivalent to $sum_{p le x} log p sim x$ meaning the quotient of the LHS and the RHS tends to $1$ as $x to infty$.
– mathworker21
Dec 28 '17 at 6:07
1
1
The integral expression is needlessly complicated. You are basically summing the log of successive primes. The prime number theorem (which is true) is equivalent to $sum_{p le x} log p sim x$ meaning the quotient of the LHS and the RHS tends to $1$ as $x to infty$.
– mathworker21
Dec 28 '17 at 6:07
The integral expression is needlessly complicated. You are basically summing the log of successive primes. The prime number theorem (which is true) is equivalent to $sum_{p le x} log p sim x$ meaning the quotient of the LHS and the RHS tends to $1$ as $x to infty$.
– mathworker21
Dec 28 '17 at 6:07
add a comment |
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The integral expression is needlessly complicated. You are basically summing the log of successive primes. The prime number theorem (which is true) is equivalent to $sum_{p le x} log p sim x$ meaning the quotient of the LHS and the RHS tends to $1$ as $x to infty$.
– mathworker21
Dec 28 '17 at 6:07