<\/p>\n\n\n\n
From @infseriesbot, prove the identity: \\(\\gamma_n = \\displaystyle\\int_1^\\infty \\frac{\\{x\\}}{x^2} (n-\\log(x)) \\log^{n-1} \\, dx \\). <\/p>\n\n\n\n
<\/p>\n\n\n\n
We have \\(\\{x\\}=x-i \\, \\text{for} \\, i \\leq x \\leq i+1\\), <\/p>\n\n\n\n
so <\/p>\n\n\n\n
\\(\\text{rhs}=\\displaystyle\\sum _{i=1}^{\\infty } \\displaystyle\\int_i^{i+1} \\frac{(x-i) (n-\\log (x)) \\log ^{n-1}(x)}{x^2} \\, dx,\\)<\/p>\n\n\n\n
and since \\(\\displaystyle\\int \\frac{(x-i) (n-x \\log ) \\left(x \\log ^{n-1}\\right)}{x^2} \\, dx=\\log ^n(x) \\left(-\\frac{i}{x}-\\frac{\\log (x)}{n+1}+1\\right)\\)<\/p>\n\n\n\n
\\(\\displaystyle \\int_i^{i+1} \\frac{(x-i) (n-\\log (x)) \\log ^{n-1}(x)}{x^2} \\, dx=\\frac{(i+1) \\log ^{n+1}(i)+(-(i+1) \\log (i+1)+n+1) \\log ^n(i+1)}{(i+1) (n+1)}\\),<\/p>\n\n\n\n
allora: <\/p>\n\n\n\n
\\(\\text{rhs}=\\underset{T\\to \\infty }{\\text{lim}}\\left(\\frac{(n-(T+1) \\log (T+1)+1) \\log ^n(T+1)}{(n+1) (T+1)}+\\sum _{i=2}^T \\frac{\\log ^n(i)}{i}\\right)\\),<\/p>\n\n\n\n
and since \\(\\frac{\\log ^n(T+1) (n-(T+1) \\log (T+1)+1)}{(n+1) (T+1)}\\overset{T}{\\to}\\frac{\\log ^{n+1}(T)}{n+1}\\),<\/p>\n\n\n\n
From the series representation of the Stieltjes Gamma function, \\(\\gamma_n\\):<\/p>\n\n\n\n
\\(\\underset{T\\to \\infty }{\\text{lim}}\\left(\\sum _{i=1}^T \\frac{\\log ^n i}{i}-\\frac{\\log ^{n+1}(T)}{n+1}\\right)=\\gamma _n\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
From @infseriesbot, prove the identity: \\(\\gamma_n = \\displaystyle\\int_1^\\infty \\frac{\\{x\\}}{x^2} (n-\\log(x)) \\log^{n-1} \\, dx \\). We have \\(\\{x\\}=x-i \\, \\text{for} \\, i \\leq x \\leq i+1\\), so \\(\\text{rhs}=\\displaystyle\\sum _{i=1}^{\\infty } \\displaystyle\\int_i^{i+1} \\frac{(x-i) (n-\\log (x)) \\log ^{n-1}(x)}{x^2} \\, dx,\\) and since \\(\\displaystyle\\int \\frac{(x-i) (n-x \\log ) \\left(x \\log ^{n-1}\\right)}{x^2} \\, dx=\\log ^n(x) \\left(-\\frac{i}{x}-\\frac{\\log (x)}{n+1}+1\\right)\\) \\(\\displaystyle \\int_i^{i+1} \\frac{(x-i) … Continue reading “A nice Limit”<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_bbp_topic_count":0,"_bbp_reply_count":0,"_bbp_total_topic_count":0,"_bbp_total_reply_count":0,"_bbp_voice_count":0,"_bbp_anonymous_reply_count":0,"_bbp_topic_count_hidden":0,"_bbp_reply_count_hidden":0,"_bbp_forum_subforum_count":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-770","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/posts\/770","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/comments?post=770"}],"version-history":[{"count":0,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/posts\/770\/revisions"}],"wp:attachment":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=770"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=770"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=770"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}