Help proving $sum_pq leq x log plog q fraclog xlog pq = sum_pq leq x log plog q + O(x)$

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I'm reading Selberg, A. (1949). An Elementary Proof of the Prime-Number Theorem After deriving his formula: $$sum_p leq x log^2 p + sum_pq leq x log p log q = 2xlog x + O(x)$$ The author states that "By partial summation we get from (formula above)" to: $$sum_p leq x log p + sum_pq leq x fraclog p log qlog pq = 2x + O(fracxlog x)$$ I don't understand how to derive this by partial sums but since I know that: $$sum_p leq x log^2 p = log x sum_p leq x log p + O(x)$$ I figure that I need the equation in question $$sum_pq leq x log plog q fraclog xlog pq = sum_pq leq x log plog q + O(x)$$ problem being that I am unable to prove this so please help. (Also if unclear note that $p$ and $q$ denote primes being not necessarily different)

sequences-and-series number-theory prime-numbers analytic-number-theory

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I'm reading Selberg, A. (1949). An Elementary Proof of the Prime-Number Theorem After deriving his formula: $$sum_p leq x log^2 p + sum_pq leq x log p log q = 2xlog x + O(x)$$ The author states that "By partial summation we get from (formula above)" to: $$sum_p leq x log p + sum_pq leq x fraclog p log qlog pq = 2x + O(fracxlog x)$$ I don't understand how to derive this by partial sums but since I know that: $$sum_p leq x log^2 p = log x sum_p leq x log p + O(x)$$ I figure that I need the equation in question $$sum_pq leq x log plog q fraclog xlog pq = sum_pq leq x log plog q + O(x)$$ problem being that I am unable to prove this so please help. (Also if unclear note that $p$ and $q$ denote primes being not necessarily different)

sequences-and-series number-theory prime-numbers analytic-number-theory

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asked 2 days ago

John

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I'm reading Selberg, A. (1949). An Elementary Proof of the Prime-Number Theorem After deriving his formula: $$sum_p leq x log^2 p + sum_pq leq x log p log q = 2xlog x + O(x)$$ The author states that "By partial summation we get from (formula above)" to: $$sum_p leq x log p + sum_pq leq x fraclog p log qlog pq = 2x + O(fracxlog x)$$ I don't understand how to derive this by partial sums but since I know that: $$sum_p leq x log^2 p = log x sum_p leq x log p + O(x)$$ I figure that I need the equation in question $$sum_pq leq x log plog q fraclog xlog pq = sum_pq leq x log plog q + O(x)$$ problem being that I am unable to prove this so please help. (Also if unclear note that $p$ and $q$ denote primes being not necessarily different)

sequences-and-series number-theory prime-numbers analytic-number-theory

share|cite|improve this question

asked 2 days ago

John

1

I'm reading Selberg, A. (1949). An Elementary Proof of the Prime-Number Theorem After deriving his formula: $$sum_p leq x log^2 p + sum_pq leq x log p log q = 2xlog x + O(x)$$ The author states that "By partial summation we get from (formula above)" to: $$sum_p leq x log p + sum_pq leq x fraclog p log qlog pq = 2x + O(fracxlog x)$$ I don't understand how to derive this by partial sums but since I know that: $$sum_p leq x log^2 p = log x sum_p leq x log p + O(x)$$ I figure that I need the equation in question $$sum_pq leq x log plog q fraclog xlog pq = sum_pq leq x log plog q + O(x)$$ problem being that I am unable to prove this so please help. (Also if unclear note that $p$ and $q$ denote primes being not necessarily different)

sequences-and-series number-theory prime-numbers analytic-number-theory

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asked 2 days ago

John

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asked 2 days ago

John

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asked 2 days ago

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If we define
beginalign
f(n) &= begincases (log n)^2 &textif n text is prime, \
quad 0 &textotherwise, endcases \
g(n) &= sum_pq = n log plog q,,
endalign
and $h(n) = f(n) + g(n)$, the formula we start with is
$$H(x) := sum_n leqslant x h(n) = 2xlog x + O(x),. tag$ast$$$
Using Abel's formula we obtain
beginalign
sum_p leqslant x log p + sum_pq leqslant x fraclog plog qlog (pq)
&= sum_2 leqslant n leqslant x frach(n)log n \
&= fracH(x)log x + int_2^x fracH(t)t(log t)^2,dt \
&= 2x + Obiggl(fracxlog xbiggr) + 2int_2^x fracdtlog t + int_2^x Obiggl(frac1(log t)^2biggr),dt \
&= 2x + 2operatornameLi(x) + Obiggl(fracxlog xbiggr) + Obiggl(fracx(log x)^2biggr) \
&= 2x + Obiggl(fracxlog xbiggr),.
endalign
The same technique gives the results for the component parts, but without the prime number theorem we cannot yet explicitly state the asymptotics of each part. Using Chebyshev's result that $vartheta(x) in Theta(x)$ we get
beginalign
F(x) &= sum_n leqslant x f(n) \
&= sum_p leqslant x (log p)cdot (log p) \
&= vartheta(x)log x - int_1^x fracvartheta(t)t,dt \
&= vartheta(x)log x + int_1^x O(1),dt \
&= vartheta(x)log x + O(x)
endalign
and, since $G(x) in O(xlog x)$ by this and $(ast)$,
beginalign
sum_n leqslant x fracg(n)log n
&= fracG(x)log x + int_2^x fracG(t)t(log t)^2,dt \
&= fracG(x)log x + Obiggl(fracxlog xbiggr),,
endalign
which is indeed
$$sum_pqleqslant x log plog qfraclog xlog (pq) = sum_pq leqslant x log plog q + O(x),.$$

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answered 6 hours ago

Daniel Fischer♦

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1 Answer
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votes

1 Answer
1

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oldest

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active

oldest

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active

oldest

votes

up vote
0
down vote

If we define
beginalign
f(n) &= begincases (log n)^2 &textif n text is prime, \
quad 0 &textotherwise, endcases \
g(n) &= sum_pq = n log plog q,,
endalign
and $h(n) = f(n) + g(n)$, the formula we start with is
$$H(x) := sum_n leqslant x h(n) = 2xlog x + O(x),. tag$ast$$$
Using Abel's formula we obtain
beginalign
sum_p leqslant x log p + sum_pq leqslant x fraclog plog qlog (pq)
&= sum_2 leqslant n leqslant x frach(n)log n \
&= fracH(x)log x + int_2^x fracH(t)t(log t)^2,dt \
&= 2x + Obiggl(fracxlog xbiggr) + 2int_2^x fracdtlog t + int_2^x Obiggl(frac1(log t)^2biggr),dt \
&= 2x + 2operatornameLi(x) + Obiggl(fracxlog xbiggr) + Obiggl(fracx(log x)^2biggr) \
&= 2x + Obiggl(fracxlog xbiggr),.
endalign
The same technique gives the results for the component parts, but without the prime number theorem we cannot yet explicitly state the asymptotics of each part. Using Chebyshev's result that $vartheta(x) in Theta(x)$ we get
beginalign
F(x) &= sum_n leqslant x f(n) \
&= sum_p leqslant x (log p)cdot (log p) \
&= vartheta(x)log x - int_1^x fracvartheta(t)t,dt \
&= vartheta(x)log x + int_1^x O(1),dt \
&= vartheta(x)log x + O(x)
endalign
and, since $G(x) in O(xlog x)$ by this and $(ast)$,
beginalign
sum_n leqslant x fracg(n)log n
&= fracG(x)log x + int_2^x fracG(t)t(log t)^2,dt \
&= fracG(x)log x + Obiggl(fracxlog xbiggr),,
endalign
which is indeed
$$sum_pqleqslant x log plog qfraclog xlog (pq) = sum_pq leqslant x log plog q + O(x),.$$

share|cite|improve this answer

answered 6 hours ago

Daniel Fischer♦

171k16153273

add a comment | 

up vote
0
down vote

If we define
beginalign
f(n) &= begincases (log n)^2 &textif n text is prime, \
quad 0 &textotherwise, endcases \
g(n) &= sum_pq = n log plog q,,
endalign
and $h(n) = f(n) + g(n)$, the formula we start with is
$$H(x) := sum_n leqslant x h(n) = 2xlog x + O(x),. tag$ast$$$
Using Abel's formula we obtain
beginalign
sum_p leqslant x log p + sum_pq leqslant x fraclog plog qlog (pq)
&= sum_2 leqslant n leqslant x frach(n)log n \
&= fracH(x)log x + int_2^x fracH(t)t(log t)^2,dt \
&= 2x + Obiggl(fracxlog xbiggr) + 2int_2^x fracdtlog t + int_2^x Obiggl(frac1(log t)^2biggr),dt \
&= 2x + 2operatornameLi(x) + Obiggl(fracxlog xbiggr) + Obiggl(fracx(log x)^2biggr) \
&= 2x + Obiggl(fracxlog xbiggr),.
endalign
The same technique gives the results for the component parts, but without the prime number theorem we cannot yet explicitly state the asymptotics of each part. Using Chebyshev's result that $vartheta(x) in Theta(x)$ we get
beginalign
F(x) &= sum_n leqslant x f(n) \
&= sum_p leqslant x (log p)cdot (log p) \
&= vartheta(x)log x - int_1^x fracvartheta(t)t,dt \
&= vartheta(x)log x + int_1^x O(1),dt \
&= vartheta(x)log x + O(x)
endalign
and, since $G(x) in O(xlog x)$ by this and $(ast)$,
beginalign
sum_n leqslant x fracg(n)log n
&= fracG(x)log x + int_2^x fracG(t)t(log t)^2,dt \
&= fracG(x)log x + Obiggl(fracxlog xbiggr),,
endalign
which is indeed
$$sum_pqleqslant x log plog qfraclog xlog (pq) = sum_pq leqslant x log plog q + O(x),.$$

share|cite|improve this answer

answered 6 hours ago

Daniel Fischer♦

171k16153273

add a comment | 

up vote
0
down vote

up vote
0
down vote

If we define
beginalign
f(n) &= begincases (log n)^2 &textif n text is prime, \
quad 0 &textotherwise, endcases \
g(n) &= sum_pq = n log plog q,,
endalign
and $h(n) = f(n) + g(n)$, the formula we start with is
$$H(x) := sum_n leqslant x h(n) = 2xlog x + O(x),. tag$ast$$$
Using Abel's formula we obtain
beginalign
sum_p leqslant x log p + sum_pq leqslant x fraclog plog qlog (pq)
&= sum_2 leqslant n leqslant x frach(n)log n \
&= fracH(x)log x + int_2^x fracH(t)t(log t)^2,dt \
&= 2x + Obiggl(fracxlog xbiggr) + 2int_2^x fracdtlog t + int_2^x Obiggl(frac1(log t)^2biggr),dt \
&= 2x + 2operatornameLi(x) + Obiggl(fracxlog xbiggr) + Obiggl(fracx(log x)^2biggr) \
&= 2x + Obiggl(fracxlog xbiggr),.
endalign
The same technique gives the results for the component parts, but without the prime number theorem we cannot yet explicitly state the asymptotics of each part. Using Chebyshev's result that $vartheta(x) in Theta(x)$ we get
beginalign
F(x) &= sum_n leqslant x f(n) \
&= sum_p leqslant x (log p)cdot (log p) \
&= vartheta(x)log x - int_1^x fracvartheta(t)t,dt \
&= vartheta(x)log x + int_1^x O(1),dt \
&= vartheta(x)log x + O(x)
endalign
and, since $G(x) in O(xlog x)$ by this and $(ast)$,
beginalign
sum_n leqslant x fracg(n)log n
&= fracG(x)log x + int_2^x fracG(t)t(log t)^2,dt \
&= fracG(x)log x + Obiggl(fracxlog xbiggr),,
endalign
which is indeed
$$sum_pqleqslant x log plog qfraclog xlog (pq) = sum_pq leqslant x log plog q + O(x),.$$

share|cite|improve this answer

answered 6 hours ago

Daniel Fischer♦

171k16153273

If we define
beginalign
f(n) &= begincases (log n)^2 &textif n text is prime, \
quad 0 &textotherwise, endcases \
g(n) &= sum_pq = n log plog q,,
endalign
and $h(n) = f(n) + g(n)$, the formula we start with is
$$H(x) := sum_n leqslant x h(n) = 2xlog x + O(x),. tag$ast$$$
Using Abel's formula we obtain
beginalign
sum_p leqslant x log p + sum_pq leqslant x fraclog plog qlog (pq)
&= sum_2 leqslant n leqslant x frach(n)log n \
&= fracH(x)log x + int_2^x fracH(t)t(log t)^2,dt \
&= 2x + Obiggl(fracxlog xbiggr) + 2int_2^x fracdtlog t + int_2^x Obiggl(frac1(log t)^2biggr),dt \
&= 2x + 2operatornameLi(x) + Obiggl(fracxlog xbiggr) + Obiggl(fracx(log x)^2biggr) \
&= 2x + Obiggl(fracxlog xbiggr),.
endalign
The same technique gives the results for the component parts, but without the prime number theorem we cannot yet explicitly state the asymptotics of each part. Using Chebyshev's result that $vartheta(x) in Theta(x)$ we get
beginalign
F(x) &= sum_n leqslant x f(n) \
&= sum_p leqslant x (log p)cdot (log p) \
&= vartheta(x)log x - int_1^x fracvartheta(t)t,dt \
&= vartheta(x)log x + int_1^x O(1),dt \
&= vartheta(x)log x + O(x)
endalign
and, since $G(x) in O(xlog x)$ by this and $(ast)$,
beginalign
sum_n leqslant x fracg(n)log n
&= fracG(x)log x + int_2^x fracG(t)t(log t)^2,dt \
&= fracG(x)log x + Obiggl(fracxlog xbiggr),,
endalign
which is indeed
$$sum_pqleqslant x log plog qfraclog xlog (pq) = sum_pq leqslant x log plog q + O(x),.$$

share|cite|improve this answer

answered 6 hours ago

Daniel Fischer♦

171k16153273

share|cite|improve this answer

share|cite|improve this answer

answered 6 hours ago

Daniel Fischer♦

171k16153273

answered 6 hours ago

Daniel Fischer♦

171k16153273

answered 6 hours ago

Daniel Fischer♦

171k16153273

171k16153273

add a comment | 

add a comment | 

 

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