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수학
상수 Mathematical Constants |
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{{{#!wiki style="margin: 0 -10px -5px; min-height: calc(1.5em + 5px)" {{{#!folding [ 펼치기 · 접기 ] {{{#!wiki style="margin: -5px -1px -11px" |
[math(^\ast)] 초월수임이 증명됨. | ||||
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[math(0)] (덧셈의 항등원) |
[math(1)] (곱셈의 항등원) |
[math(sqrt{2})] (최초로 증명된 무리수) |
[math(495)], [math(6174)] ( 카프리카 상수) |
[math(0)],
[math(1)], [math(3435)], [math(438579088)] ( 뮌하우젠 수) |
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[math(pi)] (원주율)[math(^\ast)] |
[math(tau)] (새 원주율)[math(^\ast)] |
[math(e)] (자연로그의 밑)[math(^\ast)] |
[math(varphi)] (황금수) |
[math(i)] (허수단위) |
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[math(G)] (카탈랑 상수) |
[math(zeta(3))] (아페리 상수) |
[math({rm Si}(pi))] (윌브레이엄-기브스 상수) |
[math(gamma)] (오일러-마스케로니 상수) |
[math(gamma_n)] (스틸체스 상수) |
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[math(Omega)] (오메가 상수)[math(^\ast)] |
[math(2^{sqrt{2}})] (겔폰트-슈나이더 상수)[math(^\ast)] |
[math(C_n,)] (챔퍼나운 상수)[math(^\ast)] |
[math(A,)] (글레이셔-킨켈린 상수) |
[math(A_k,)] (벤더스키-아담칙 상수) |
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[math(-e, {rm Ei}(-1))] (곰페르츠 상수) |
[math(mu)] (라마누잔-졸트너 상수) |
[math(B_{2})], [math(B_{4})] (브룬 상수) |
[math(rho)] (플라스틱 상수) |
[math(delta)], [math(alpha)] (파이겐바움 상수) |
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1. 개요
Stieltjes constants스틸체스 상수는 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 성질 문단에서 볼 수 있다.[1][2]
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[math(\displaystyle \begin{aligned}
\zeta(s) &= \frac1{s-1} +\sum_{n=0}^\infty \frac{(-1)^n}{n!} \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) (s-1)^n \\ &= \frac1{s-1} +\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n (s-1)^n \end{aligned} )] |
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[math(\displaystyle \begin{aligned}
\gamma_n &= \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) \\ &= \lim_{m\to\infty} \Biggl( \sum_{k=1}^m \frac{(\ln k)^n}k -\frac{(\ln m)^{n+1}}{n+1} \Biggr) \end{aligned} )] |
2. 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 절댓값이 대체로 작아지는 것을 알 수 있다. 초반 이후로 큰 [math(n)]값에 대해서는 스틸체스 상수의 절댓값이 급격히 커진다. 위의 표에 없는 다른 [math(n)]에 대한 스틸체스 상수의 값은 이 링크에서 볼 수 있다. 표에 나오지 않은 소수점 뒷자리를 좀 더 보고 싶으면 링크로 들어가서 해당 수를 클릭하면 된다. [math(\gamma_n)]의 그래프는 여기[3]에서 볼 수 있다.
3. 항등식
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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 함수이다. 그러면 다음이 성립한다.
{{{#!wiki style="text-align: center"
[math(\displaystyle \begin{aligned}
\int_0^1 \!\left\{ \frac1x \right\} \ln x \,{\rm d}x &= \gamma+\gamma_1-1 \\
&\approx -0.4956001806
\end{aligned} )]}}}
4단계로 나누어서 증명한다. 적분 상수는 [math({\sf const.})]라고 표기하였다. 증명 과정에서 나오는 [math(H_n)]은 조화수이다.
[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} )]
}}} ||
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위 식을 좀 더 일반화하면 다음과 같다. [math(0)] 이상의 정수 [math(n)]에 대해 다음이 성립한다. 여기서 [math(n^{\underline i})]은
하강 계승(=
순열)이다.
\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} )]}}}||
[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}. \\
\Rightarrow 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} )]
{\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}. \\
\Rightarrow 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} )]
[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} )]
[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} )]
[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 때에 대한 예시는 아래와 같다. [math(n=0)]일 때에 대해서는
오일러-마스케로니 상수 문서의
항등식 문단을 참고하라.
[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} )]}}}||