[include(틀:토막글)]
[목차]

[include(틀:수학상수의 목록)]

== 개요 ==
{{{+1 Stieltjes constants}}}

스틸체스 상수는 [[제타 함수|리만 제타 함수 [math(\zeta(s))]]]를 [math(s=1)]을 기준으로 로랑 급수 전개를 했을 때 볼 수 있는 상수로, 다음 식의 [math(\gamma_n)]에 해당한다. 유도 과정은 [[제타 함수]] 문서의 [[제타 함수#성질|성질]] 문단에서 볼 수 있다.[* [[https://freshrimpsushi.github.io/posts/derivation-of-laurent-expansion-of-riemann-zeta-function/|다른 유도 과정]]][* [[리만 가설]]을 만든 그 리만이다.]
||<bgcolor=#fff,#1f2023><table width=100%><tablebordercolor=#fff,#1f2023><:>{{{#!wiki style="text-align: center"
[math(\displaystyle
\zeta(s) = \frac1{s-1} +\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n(s-1)^n
)]}}}||
스틸체스 상수는 다음과 같이 극한을 이용해 표현할 수 있다.
||<bgcolor=#fff,#1f2023><table width=100%><tablebordercolor=#fff,#1f2023><:>{{{#!wiki style="text-align: center"
[math(\displaystyle \begin{aligned}
\gamma_n &= \lim_{m\to\infty} \Biggl( \sum_{k=1}^m \frac{(\ln k)^n}k -\frac{(\ln m)^{n+1}}{n+1} \Biggr) \\
&= \lim_{m\to\infty} \Biggl( \sum_{k=1}^m \frac{(\ln k)^n}k -\int_1^m \frac{(\ln x)^n}x \,{\rm d}x \Biggr)
\end{aligned} )]}}}||
특히 [math(\gamma_0=\gamma\approx0.5772156649)]는 [[오일러-마스케로니 상수]]이다.

== 값 ==
||[math(n)]||[math(\gamma_n)]의 근삿값||
||[math(0)]||[math(+0.5772156649015328606065120900824024310421593359)]||
||[math(1)]||[math(-0.0728158454836767248605863758749013191377363383)]||
||[math(2)]||[math(-0.0096903631928723184845303860352125293590658061)]||
||[math(3)]||[math(+0.0020538344203033458661600465427533842857158044)]||
||[math(4)]||[math(+0.0023253700654673000574681701775260680009044694)]||
||[math(5)]||[math(+0.0007933238173010627017533348774444448307315394)]||
||[math(6)]||[math(-0.0002387693454301996098724218419080042777837151)]||
||[math(7)]||[math(-0.0005272895670577510460740975054788582819962534)]||
||[math(8)]||[math(-0.0003521233538030395096020521650012087417291805)]||
||[math(9)]||[math(-0.0000343947744180880481779146237982273906207895)]||
||[math(10)]||[math(+0.0002053328149090647946837222892370653029598537)]||
||[math(100)]||[math(-4.2534015717080269623144385197278358247028931053\times10^{17})]||
||[math(1000)]||[math(-1.5709538442047449345494023425120825242380299554\times10^{486})]||
||[math(10000)]||[math(-2.2104970567221060862971082857536501900234397174\times10^{6883})]||
||[math(100000)]||[math(+1.9919273063125410956582272431568589205211659777\times10^{83432})]||

초반의 작은 [math(n)]에 대해서는, [math(n)]이 커질수록 [[절댓값]]이 대체로 작아지는 것을 알 수 있다. 초반 이후로 큰 [math(n)]값에 대해서는 스틸체스 상수의 [[절댓값]]이 급격히 커진다. 위의 표에 없는 다른 [math(n)]에 대한 스틸체스 상수의 값은 [[https://beta.lmfdb.org/riemann/stieltjes/|이 링크]]에서 볼 수 있다. 표에 나오지 않은 소수점 뒷자리를 좀 더 보고 싶으면 링크로 들어가서 해당 수를 클릭하면 된다. [math(\gamma_n)]의 그래프는 [[https://www.semanticscholar.org/paper/A-BOUND-FOR-FRACTIONAL-STIELTJES-CONSTANTS-Farr-Pauli/24ebe68d1116c979171073f6f9f8ae420e79df0e/figure/0|여기]]에서 볼 수 있다.[* 이 그래프에 나와 있는 파란 선 위의 빨간 점이 [math(\gamma_n)]의 값이다. 파란 선은 스틸체스 상수 [math(\gamma_n)]에서 [math(n)]의 값을 실수로 확장한 "fractional Stieltjes constants"의 그래프이다.]

== 항등식 ==
 * [[최대 정수 함수#s-2|소수 부분을 나타내는 함수]] [math(\left\{x\right\})]를 정의하자. 즉, [math(\left\{x\right\} = x - \left\lfloor x \right\rfloor)]라고 하자. 여기서 [math(\left\lfloor x \right\rfloor)]는 [[최대 정수 함수]]이다. 그러면 다음이 성립한다.
 {{{#!wiki style="text-align: center"
[br][math(\displaystyle \begin{aligned}
\int_0^1 \!\left\{ \frac1x \right\} \ln x \,{\rm d}x &= \gamma+\gamma_1-1 \\
&\approx -0.4956001806
\end{aligned} )]}}}
 ||<table width=100%>{{{#!folding [증명]
----
4단계로 나누어서 증명한다. 적분 상수는 [math({\sf const.})]라고 표기하였다. 증명 과정에서 나오는 [math(H_n)]은 [[조화수(수학)|조화수]]이다.
{{{#!wiki style="text-align: center"
[br][math(\displaystyle \begin{aligned}
\int \frac{\ln u}{u^2} \,{\rm d}u &= \ln u \cdot \biggl( -\frac1u \biggr) -\int \frac1u \cdot \biggl( -\frac1u \biggr) {\rm d}u = -\frac{\ln u}u -\frac1u +{\sf const.} \\

\int_k^{k+1} \frac{k\ln u}{u^2} \,{\rm d}u &= \biggl[ -\frac{k\ln u}u -\frac ku \biggr]_k^{k+1} = -\frac{k\ln(k+1)}{k+1} -\frac k{k+1} +\ln k +1 \\
&= \ln k +1 -\frac k{k+1} -\biggl( 1-\frac1{k+1} \biggr) \!\ln(k+1) \\
&= \ln k +\frac1{k+1} -\ln(k+1) +\frac{\ln(k+1)}{k+1} \\

\sum_{k=1}^{n-1} \int_k^{k+1} \frac{k\ln u}{u^2} \,{\rm d}u &= \sum_{k=1}^{n-1} \biggl( \ln k -\ln(k+1) +\frac1{k+1} +\frac{\ln(k+1)}{k+1} \biggr) \\
&= \sum_{k=1}^{n-1} \Bigl( \ln k -\ln(k+1) \Bigr) +\sum_{k=1}^{n-1} \frac1{k+1} +\sum_{k=1}^{n-1} \frac{\ln(k+1)}{k+1} \\
&= -\ln n +(H_n-1) +\sum_{k=2}^n \frac{\ln k}k \\

\therefore \int_0^1 \!\left\{ \frac1x \right\} \ln x \,{\rm d}x &= \int_\infty^1 \{u\} \ln{\frac1u} \,\biggl( -\frac{{\rm d}u}{u^2} \biggr) = \int_1^\infty \cfrac{\left\lfloor u \right\rfloor -u}{u^2} \,\ln u \,{\rm d}u \\
&= \lim_{n\to\infty} \int_1^n \biggl( \frac{\left\lfloor u \right\rfloor \ln u}{u^2} -\frac{\ln u}u \biggr) {\rm d}u \\
&= \lim_{n\to\infty} \Biggl( \sum_{k=1}^{n-1} \int_k^{k+1} \frac{\left\lfloor u \right\rfloor \ln u}{u^2} \,{\rm d}u -\int_1^n \frac{\ln u}u \,{\rm d}u \Biggr) \\
&= \lim_{n\to\infty} \Biggl( \sum_{k=1}^{n-1} \int_k^{k+1} \frac{k\ln u}{u^2} \,{\rm d}u -\biggl[ \frac12 \ln^2u \biggr]_1^n \Biggr) \\
&= \lim_{n\to\infty} \Biggl( -\ln n +H_n -1 +\sum_{k=2}^n \frac{\ln k}k -\frac12 \ln^2n \Biggr) \\
&= \lim_{n\to\infty} \Biggl( (H_n -\ln n) + \Biggl( \sum_{k=1}^n \frac{\ln k}k -\frac12 \ln^2n \Biggr) -1 \Biggr) \\
&= \gamma +\gamma_1 -1
\end{aligned} )]}}}
}}} ||

 * 위 식을 좀 더 일반화하면 다음과 같다. [math(0)] 이상의 정수 [math(n)]에 대해 다음이 성립한다. 여기서 [math(n^{\underline i})]은 [[계승(수학)#s-2|하강 계승]](= [[순열]])이다.
 ||<bgcolor=#fff,#1f2023><table width=100%><tablebordercolor=#fff,#1f2023><:>{{{#!wiki style="text-align: center"
[math(\displaystyle \begin{aligned}
\int_0^1 \!\left\{ \frac1x \right\} \ln^n x \,{\rm d}x &= (-1)^{n+1} \Biggl( \sum_{i=0}^n n^{\underline i} \gamma_{n-i} -n! \Biggr) \\
&= (-1)^{n+1} \biggl( \frac{n!}{n!}\gamma_n +\frac{n!}{(n-1)!}\gamma_{n-1} +\frac{n!}{(n-2)!}\gamma_{n-2} +\cdots +\frac{n!}{3!}\gamma_3 +\frac{n!}{2!}\gamma_2 +\frac{n!}{1!}\gamma_1 +\frac{n!}{0!}\gamma -n! \biggr) \\
&= (-1)^{n+1} \biggl( \gamma_n +n\gamma_{n-1} +n(n-1)\gamma_{n-2} +\cdots +\frac{n!}6\gamma_3 +\frac{n!}2\gamma_2 +n!\cdot\gamma_1 +n!\cdot\gamma -n! \biggr)
\end{aligned} )]}}}||
 
 ||<table width=100%>{{{#!folding [증명]
----
{{{#!wiki style="text-align: center"
[math(\displaystyle \begin{aligned}
{\sf Let}: I_n &= \int \frac{\ln^nu}{u^2} \,{\rm d}u \\
\Rightarrow I_0 &= \int \frac1{u^2} \,{\rm d}u = -\frac1u + {\sf const}. \\
\therefore I_n &= \int \frac{\ln^nu}{u^2} \,{\rm d}u = \int \ln^nu \cdot \frac1{u^2} \,{\rm d}u \\
&= \ln^nu \cdot \!\left( -\frac1u \right) -\int n\ln^{n-1}u \cdot \frac1u \cdot \!\left( -\frac1u \right) \!{\rm d}u \\
&= -\frac{\ln^nu}u +n\int \frac{\ln^{n-1}u}{u^2} \,{\rm d}u \\
&= -\frac{\ln^nu}u +nI_{n-1} \\
&= -\frac{\ln^nu}u -\frac{n\ln^{n-1}u}u +n(n-1)I_{n-2} \\
&= \cdots \\
&= -\frac{\ln^nu}u -\frac{n\ln^{n-1}u}u -\cdots -\frac{n(n-1)\cdots3\cdot2\cdot\ln u}u +n(n-1)\cdots2\cdot1\cdot I_0 \\
&= -\frac{\ln^nu}u -\frac{n\ln^{n-1}u}u -\cdots -\frac{n(n-1)\cdots3\cdot2\cdot\ln u}u -\frac{n!}u +{\sf const}. \\
&= -\sum_{i=0}^n \frac{n^{\underline i} \ln^{n-i}u}u +{\sf const}. \\
\therefore \int \frac{\ln^nu}{u^2} \,{\rm d}u &= -\sum_{i=0}^n \frac{n^{\underline i} \ln^{n-i}u}u +{\sf const}.
\end{aligned} )]}}}
{{{#!wiki style="text-align: center"
[br][math(\displaystyle \begin{aligned}
\int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u &= \!\Biggl[ -\sum_{i=0}^n \frac{kn^{\underline i} \ln^{n-i}u}u \Biggr]_k^{k+1} \\
&= -\sum_{i=0}^n \frac{kn^{\underline i} \ln^{n-i} (k+1)}{k+1} +\sum_{i=0}^n n^{\underline i} \ln^{n-i}k \\
&= -\frac k{k+1} \sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) +\sum_{i=0}^n n^{\underline i} \ln^{n-i}k \\
&= -\biggl( 1-\frac1{k+1} \biggr) \!\sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) +\sum_{i=0}^n n^{\underline i} \ln^{n-i}k \\
&= \frac1{k+1} \Biggl( \sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) \Biggr) \!+\!\Biggl( \sum_{i=0}^n n^{\underline i} \ln^{n-i}k \Biggr) \!-\!\Biggl( \sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) \Biggr) \\
&= \frac1{k+1} \Biggl( {\color{DeepSkyBlue} \ln^n (k+1) } +\sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} n! } \Biggr) \\
&\quad +\!\Biggl( \sum_{i=0}^{n-2} n^{\underline i} \ln^{n-i}k +{\color{limegreen} n^{\underline{n-1}} \ln k } +n! \Biggr) \!-\!\Biggl( \sum_{i=0}^{n-2} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} n^{\underline{n-1}} \ln (k+1) } +n! \Biggr) \\
&= {\color{DeepSkyBlue} \frac{\ln^n (k+1)}{k+1} } +\frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \frac{n!}{k+1} } \\
&\quad +\sum_{i=0}^{n-2} n^{\underline i} (\ln^{n-i}k -\ln^{n-i} (k+1)) +{\color{limegreen} n^{\underline{n-1}} (\ln k -\ln (k+1)) } \\
&= {\color{DeepSkyBlue} \frac{\ln^n (k+1)}{k+1} } +\frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \frac{n!}{k+1} } \\
&\quad +\sum_{i=1}^{n-1} n^{\underline{i-1}} (\ln^{n-i+1}k -\ln^{n-i+1} (k+1)) +{\color{limegreen} n^{\underline{n-1}} (\ln k -\ln (k+1)) }
\end{aligned} )]}}}
{{{#!wiki style="text-align: center"
[br][math(\displaystyle \begin{aligned}
\sum_{k=1}^{m-1} &\int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u {\color{DeepSkyBlue} \,-\,\frac{\ln^{n+1}m}{n+1} } \\
&= {\color{DeepSkyBlue} -\frac{\ln^{n+1}m}{n+1} } +\sum_{k=1}^{m-1} \Biggl( {\color{DeepSkyBlue} \frac{\ln^n (k+1)}{k+1} } +\frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \frac{n!}{k+1} } \\
&\quad +\sum_{i=1}^{n-1} n^{\underline{i-1}} (\ln^{n-i+1}k -\ln^{n-i+1} (k+1)) +{\color{limegreen} n^{\underline{n-1}} (\ln k -\ln (k+1)) } \Biggr) \\
&= {\color{DeepSkyBlue} -\frac{\ln^{n+1}m}{n+1} +\sum_{k=1}^{m-1} \frac{\ln^n (k+1)}{k+1} } +\sum_{k=1}^{m-1} \frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \sum_{k=1}^{m-1} \frac{n!}{k+1} } \\
&\quad +\sum_{k=1}^{m-1} \sum_{i=1}^{n-1} n^{\underline{i-1}} (\ln^{n+1-i}k -\ln^{n+1-i} (k+1)) +{\color{limegreen} \sum_{k=1}^{m-1} n^{\underline{n-1}} (\ln k -\ln (k+1)) } \\
&= {\color{DeepSkyBlue} -\frac{\ln^{n+1}m}{n+1} +\sum_{k=2}^m \frac{\ln^n k}k } +\sum_{k=2}^m \frac1k \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i}k +{\color{limegreen} \sum_{k=2}^m \frac{n!}k } \\
&\quad +\sum_{i=1}^{n-1} n^{\underline{i-1}} \sum_{k=1}^{m-1} (\ln^{n+1-i}k -\ln^{n+1-i} (k+1)) {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \sum_{k=2}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} } +\sum_{k=2}^m \frac1k \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i}k +{\color{limegreen} \sum_{k=2}^m \frac{n!}k } +\sum_{i=1}^{n-1} n^{\underline{i-1}} (-\ln^{n+1-i}m) {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} } +\sum_{k=1}^m \frac1k \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i}k +{\color{limegreen} \sum_{k=1}^m \frac{n!}k -n! } -\sum_{i=1}^{n-1} n^{\underline{i-1}} \ln^{n+1-i}m {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} } +\sum_{i=1}^{n-1} \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k +{\color{limegreen} \sum_{k=1}^m \frac{n!}k -n! } -\sum_{i=1}^{n-1} n^{\underline{i-1}} \ln^{n-i+1}m {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \!\Biggl( \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} \Biggr) } \!+\sum_{i=1}^{n-1} \Biggl( \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k -n^{\underline{i-1}} \ln^{n-i+1}m \Biggr) \!+{\color{limegreen} \!\Biggl( \sum_{k=1}^m \frac{n!}k -n^{\underline{n-1}} \ln m \Biggr) \!-n! } \\
&\qquad {\sf Note\ 1}: n^{\underline{i-1}} = \frac{n!}{(n-(i-1))!} = \frac{n!}{(n-i+1)!} = \frac{n!}{(n-i)!}\cdot\frac1{n-i+1} = \frac{n^{\underline i}}{n-i+1} \\
&\qquad {\sf Note\ 2}: n^{\underline{n-1}} = \frac{n!}{1!} = \frac{n!}{0!\cdot1} = \frac{n^{\underline n}}1 \\
&= \!\Biggl( \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} \Biggr) \!+\sum_{i=1}^{n-1} \Biggl( \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k -\frac{n^{\underline i} \ln^{n-i+1}m}{n-i+1} \Biggr) \!+\!\Biggl( \sum_{k=1}^m \frac{n!}k -\frac{n^{\underline n} \ln m}1 \Biggr) \!-n! \\
&= \sum_{i=0}^n \Biggl( \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k -\frac{n^{\underline i} \ln^{n-i+1}m}{n-i+1} \Biggr) \!-n! \\
&= \sum_{i=0}^n n^{\underline i} \Biggl( \sum_{k=1}^m \frac{\ln^{n-i}k}k -\frac{\ln^{n-i+1}m}{n-i+1} \Biggr) \!-n!
\end{aligned} )]}}}
{{{#!wiki style="text-align: center"
[br][math(\displaystyle \begin{aligned}
\therefore \int_0^1 \!\left\{ \frac1x \right\} \ln^nx \,{\rm d}x &= \int_\infty^1 \{u\} \ln^n u^{-1} \biggl( \!-\frac{{\rm d}u}{u^2} \biggr) \\
&= (-1)^{n} \int_1^\infty \frac{u-\!\left\lfloor u \right\rfloor}{u^2} \ln^nu \,{\rm d}u \\
&= (-1)^{n+1} \int_1^\infty \frac{\left\lfloor u \right\rfloor \!-u}{u^2} \ln^nu \,{\rm d}u \\
&= (-1)^{n+1} \lim_{m\to\infty} \int_1^m \biggl( \frac{\left\lfloor u \right\rfloor \!\ln^nu}{u^2} -\frac{\ln^nu}u \biggr) {\rm d}u \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{k=1}^{m-1} \int_k^{k+1} \frac{\left\lfloor u \right\rfloor \!\ln^nu}{u^2} \,{\rm d}u -\int_1^m \frac{\ln^nu}u \,{\rm d}u \Biggr) \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{k=1}^{m-1} \int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u -\biggl[ \frac{\ln^{n+1} u}{n+1} \biggr]_1^m \Biggr) \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{k=1}^{m-1} \int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u -\frac{\ln^{n+1} m}{n+1} \Biggr) \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{i=0}^n n^{\underline i} \Biggl( \sum_{k=1}^m \frac{\ln^{n-i}k}k -\frac{\ln^{n-i+1}m}{n-i+1} \Biggr) \!-n! \Biggr) \\
&= (-1)^{n+1} \Biggl( \sum_{i=0}^n n^{\underline i} \lim_{m\to\infty} \Biggl( \sum_{k=1}^m \frac{\ln^{n-i}k}k -\frac{\ln^{n-i+1}m}{n-i+1} \Biggr) \!-n! \Biggr) \\
&= (-1)^{n+1} \Biggl( \sum_{i=0}^n n^{\underline i} \gamma_{n-i} -n! \Biggr) \\
\end{aligned} )]}}}
}}} ||
 [math(n=5)]일 때의 구체적 예시는 아래와 같다.
 ||<bgcolor=#fff,#1f2023><table width=100%><tablebordercolor=#fff,#1f2023><:>{{{#!wiki style="text-align: center"
[math(\displaystyle \begin{aligned}
\int_0^1 \!\left\{ \frac1x \right\} \ln^5x \,{\rm d}x &= (-1)^{5+1} (\gamma_5 +5\cdot\gamma_4 +5\cdot4\cdot\gamma_3 +5\cdot4\cdot3\cdot\gamma_2 +5\cdot4\cdot3\cdot2\cdot\gamma_1 +5\cdot4\cdot3\cdot2\cdot1\cdot\gamma -5!) \\
&= \gamma_5 +5\gamma_4 +20\gamma_3 +60\gamma_2 +120\gamma_1 +120\gamma -120 \\
&\approx -59.9999465989
\end{aligned} )]}}}||

== 관련 문서 ==
 * [[오일러-마스케로니 상수]]
 * [[제타 함수]]

[[분류:해석학(수학)]][[분류:수학 용어]][[분류:무리수]][[분류:수학상수]]