{"id":660,"date":"2014-01-20T01:28:44","date_gmt":"2014-01-20T06:28:44","guid":{"rendered":"http:\/\/www.curtisbright.com\/bln\/?p=660"},"modified":"2014-01-20T01:28:44","modified_gmt":"2014-01-20T06:28:44","slug":"double-logarithm-summation-over-primes","status":"publish","type":"post","link":"http:\/\/localhost\/blog\/index.php\/2014\/01\/20\/double-logarithm-summation-over-primes\/","title":{"rendered":"Double logarithm summation over primes"},"content":{"rendered":"

It is well known1<\/a><\/sup>\u00a0that<\/p>\n

\\[ \\sum_{p\\leq x}\\ln p = x + O\\biggl(\\frac{x}{e^{c\\sqrt{\\ln x}}}\\biggr) \\]<\/p>\n

where $c>0$ is a constant and the summation runs over the primes. In fact, under the Riemann hypothesis, one even has<\/p>\n

\\[ \\sum_{p\\leq x}\\ln p = x + O(x^{1\/2+\\epsilon}) \\]<\/p>\n

for any $\\epsilon>0$. Since $e^{c\\sqrt{\\ln x}}$ grows slower than any power of $x$, the second statement gives a better approximation.<\/p>\n

A related question, but one I wasn’t familiar with, is to give a similar asymptotic result for the summation with $\\ln p$ replaced by $\\ln\\ln p$. In other words, to estimate the quantity<\/p>\n

\\[ \\sum_{p\\leq x}\\ln\\ln p .\u00a0\\]<\/p>\n

To do this, we may employ Abel’s summation formula<\/a>\u00a0with<\/p>\n

\\[ a_n := \\begin{cases}
\n1 & \\text{if $n$ is prime} \\\\
\n0 & \\text{otherwise}
\n\\end{cases} \\]<\/p>\n

and $\\phi(n):=\\ln\\ln n$. Then we have<\/p>\n

\\[ \\sum_{p\\leq x}\\ln\\ln p = \\pi(x)\\ln\\ln x-\\int_2^x\\frac{\\pi(t)}{t\\ln t}\\mathrm{d}t . \\]<\/p>\n

By the prime number theorem<\/a>\u00a0we have $\\pi(t)=t\/\\ln t+O(t\/\\ln(t)^2)$, so<\/p>\n

\\[ \\int_2^x\\frac{\\pi(t)}{t\\ln t}\\mathrm{d}t = \\int_2^x\\frac{\\mathrm{d}t}{\\ln(t)^2}+O\\biggl(\\int_2^x\\frac{\\mathrm{d}t}{\\ln(t)^3}\\biggr) . \\]<\/p>\n

By Wolfram Alpha<\/a>\u00a0we have that<\/p>\n

\\[ \\int_2^x\\frac{\\mathrm{d}t}{\\ln(t)^2} = \\DeclareMathOperator{\\li}{li}\\li(x)-\\frac{x}{\\ln x} + O(1) = \\frac{x}{\\ln(x)^2} + O\\biggl(\\frac{x}{\\ln(x)^3}\\biggr) ,\u00a0\\]<\/p>\n

with the latter equality following from the asymptotic expansion of the logarithmic integral<\/a>.<\/p>\n

It remains to estimate the integral $\\int_2^x\\ln(t)^{-3}\\,\\mathrm{d}t$. Actually, this is not entirely straightforward, but a trick is to split the integral into two (around $\\sqrt{x}$) and then estimate each, as follows:<\/p>\n

\\[ \\int_2^{\\sqrt{x}}\\frac{\\mathrm{d}t}{\\ln(t)^3} + \\int_{\\sqrt{x}}^x\\frac{\\mathrm{d}t}{\\ln(t)^3} \\leq \\frac{\\sqrt{x}-2}{\\ln(2)^3} + \\frac{x-\\sqrt{x}}{\\ln(\\sqrt{x})^3} = O\\biggl(\\frac{x}{\\ln(x)^3}\\biggr)\u00a0\\]<\/p>\n

Putting everything together, we find the result<\/p>\n

\\[ \\sum_{p\\leq x}\\ln\\ln p =\\pi(x)\\ln\\ln x-\\frac{x}{\\ln(x)^2} +O\\biggl(\\frac{x}{\\ln(x)^3}\\biggr) .\u00a0\\]<\/p>\n

\n
<\/div>\n
    \n
  1. For example, see (2.29) in\u00a0Approximate formulas for some functions of prime numbers<\/em><\/a> by Rosser and Schoenfeld. ↩<\/a><\/span><\/li>\n<\/ol>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    It is well known1\u00a0that \\[ \\sum_{p\\leq x}\\ln p = x + O\\biggl(\\frac{x}{e^{c\\sqrt{\\ln x}}}\\biggr) \\] where $c>0$ is a constant and the summation runs over the primes. In fact, under the Riemann hypothesis, one even has \\[ \\sum_{p\\leq x}\\ln p = … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"_links":{"self":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/660"}],"collection":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/comments?post=660"}],"version-history":[{"count":0,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/660\/revisions"}],"wp:attachment":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=660"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=660"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=660"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}