폴리로그함수

역사 raw
대문 랜덤 문서 최근 토론
분류

1. 개요2. 미적분3. 성질4. 알려진 함숫값5. 관련 문서

1. 개요[편집]

폴리로그함수(Polylogarithm) 혹은 다중로그는 특수함수의 하나로, 로그함수를 일반화한 이변수 함수이다. 표기는 Lis(x)\mathrm{Li}_s(x)로 하며, 정의는 다음과 같다.

Lis(x)≡∑n=1∞xnns\displaystyle \mathrm{Li}_s ( x ) \equiv \sum_{n=1}^{\infty} \frac{x^n}{n^s}

위 급수는 임의의 복소수 ss와 ∣x∣<1|x|<1인 복소수 xx에 대해 수렴함이 알려져 있다. 또한, s=1s=1인 경우 로그함수 꼴로 표현된다. 즉,

Li1(x)=−ln⁡(1−x)\displaystyle \mathrm{Li}_1(x)=-\ln{(1-x)}

이다. 로그 함수의 매클로린 급수를 사용해서 증명할 수 있다. 폴리로그함수는 해석적 연속에 의해 ∣x∣≥1|x|\ge1인 복소수 xx에 대해서도 잘 정의된다. 자세한 내용은 영문 위키피디아 문서를 참고하라.

한편, Li2(x)\mathrm{Li}_2(x)와 Li3(x)\mathrm{Li}_3(x)는 자주 쓰이는 관계로 각각 이중로그(Dilogarithm), 삼중로그(Trilogarithm)라고도 부른다. 특히 Li2(x)\mathrm{Li}_2(x)는 Spence's Function이라고도 불린다.

몇몇 폴리로그함수에 대한 그래프는 아래와 같다.

파일:나무_폴리로그_그래프_NeW.png

무한급수의 형태에서 알 수 있듯 리만 제타 함수 ζ(x)\zeta(x)와 관련이 있다.[1]

2. 미적분[편집]

  • 폴리로그함수는 다음과 같이 적분을 통해 재귀적으로도 정의된다.

    Lis+1(x)=∫0xLis(t)t dt=∫01Lis(xt)t dt\displaystyle \mathrm{Li}_{s+1}(x)=\int_0^x\frac{\mathrm{Li}_s(t)}t\,\mathrm{d}t=\int_0^1\frac{\mathrm{Li}_s(xt)}t\,\mathrm{d}t
    [증명]

    먼저 xt=yxt=y로 치환함으로써 시작한다. (xt=y⇒x dt=dy)(xt=y\Rightarrow x\,\mathrm{d}t=\mathrm{d}y)

    ∫01Lis(xt)t dt=∫0xLis(y)y/xdyx=∫0xLis(y)y dy=∫0xLis(t)t dt\displaystyle \begin{aligned} \int_0^1\frac{\mathrm{Li}_s(xt)}t\,\mathrm{d}t&=\int_0^x\frac{\mathrm{Li}_s(y)}{y/x}\frac{\mathrm{d}y}{x} \\ &=\int_0^x\frac{\mathrm{Li}_s(y)}y\,\mathrm{d}y \\ &=\int_0^x\frac{\mathrm{Li}_s(t)}t\,\mathrm{d}t \\ \end{aligned}


    이어서 무한급수꼴 정의를 사용하면 된다.

    ∫0xLis(t)t dt=∫0x1t∑n=1∞tnns dt=∑n=1∞1ns∫0xtn−1 dt=∑n=1∞1ns⋅1nxn=∑n=1∞xnns+1=Lis+1(x)\displaystyle \begin{aligned} \int_0^x\frac{\mathrm{Li}_s(t)}t\,\mathrm{d}t&=\int_0^x\frac1t\sum_{n=1}^{\infty}\frac{t^n}{n^s}\,\mathrm{d}t \\ &=\sum_{n=1}^{\infty}\frac1{n^s}\int_0^xt^{n-1}\,\mathrm{d}t \\ &=\sum_{n=1}^{\infty}\frac1{n^s}\cdot\frac1nx^n \\ &=\sum_{n=1}^{\infty}\frac{x^n}{n^{s+1}} \\ &=\mathrm{Li}_{s+1}(x) \end{aligned}
    특히 Li2(x)\mathrm{Li}_2(x)의 경우, 위의 적분식과 Li1(x)=−ln⁡(1−x)\displaystyle \mathrm{Li}_1(x)=-\ln{(1-x)}임을 사용해서 다음과 같이 표현할 수 있다.

    Li2(x)=∫0x−ln⁡(1−t)t dt=∫01−ln⁡(1−xt)t dt\displaystyle \mathrm{Li}_2(x)=\int_0^x\frac{-\ln{(1-t)}}t\,\mathrm{d}t=\int_0^1\frac{-\ln{(1-xt)}}t\,\mathrm{d}t
  • 위의 적분 관계식을 사용하면 폴리로그함수의 미분을 구할 수 있다.

    ddxLis(x)=Lis−1(x)x\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_s(x)=\frac{\mathrm{Li}_{s-1}(x)}{x}

    폴리로그함수와 임의의 함수 f(x)f(x)가 합성된 경우에도 위의 적분 관계식을 사용해 미분 가능하다. 이 때는 정적분으로 정의된 함수를 미분하는 방법을 사용하면 된다.

    ddxLis(f(x))=ddx∫0f(x)Lis−1(t)t dt=Lis−1(f(x))f(x)⋅f′(x)\displaystyle \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{Li}_s(f(x)) &= \frac{\mathrm{d}}{\mathrm{d}x} \int_0^{f(x)} \frac{\mathrm{Li}_{s-1}(t)}t \,\mathrm{d}t \\ &= \frac{\mathrm{Li}_{s-1}(f(x))}{f(x)} \cdot f'(x) \end{aligned}

    이를 사용하면 상수 a∈Ra\in\mathbb{R}에 대해 다음 식도 얻을 수 있다.

    ddxLis(ax)=Lis−1(ax)x\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{Li}_s(ax) = \frac{\mathrm{Li}_{s-1}(ax)}{x}
  • 한편, xx가 아니라 ss로 미분할 수도 있다.

    ddsLis(x)=−∑n=1∞xnln⁡nns\displaystyle \frac{\mathrm{d}}{\mathrm{d}s}\mathrm{Li}_s(x)=-\sum_{n=1}^{\infty}\frac{x^n\ln n}{n^s}
    [증명]

    ddsLis(x)=dds∑n=1∞xnns=∑n=1∞xnddsn−s=∑n=1∞xn⋅(−n−sln⁡n)=−∑n=1∞xnln⁡nns\displaystyle \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}s}\mathrm{Li}_s(x)&=\frac{\mathrm{d}}{\mathrm{d}s}\sum_{n=1}^{\infty}\frac{x^n}{n^s} \\ &=\sum_{n=1}^{\infty}x^n\frac{\mathrm{d}}{\mathrm{d}s}n^{-s} \\ &=\sum_{n=1}^{\infty}x^n\cdot(-n^{-s}\ln n) \\ &=-\sum_{n=1}^{\infty}\frac{x^n\ln n}{n^s} \end{aligned}

3. 성질[편집]

  • Lis(1)=∑n=1∞1ns=ζ(s)\displaystyle \mathrm{Li}_s(1)=\sum_{n=1}^{\infty}\frac1{n^s}=\zeta(s)
  • Lis(−1)=∑n=1∞(−1)nns=(21−s−1)ζ(s)\displaystyle \mathrm{Li}_s(-1)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}=(2^{1-s}-1)\zeta(s)
  • Duplication Formula

    Lis(x)+Lis(−x)=21−sLis(x2)\displaystyle \mathrm{Li}_s(x)+\mathrm{Li}_s(-x)=2^{1-s}\mathrm{Li}_s(x^2)
    [증명]

    Lis(x)+Lis(−x)=∑n=1∞xnns+∑n=1∞(−x)nns=2∑n=2,4,⋯∞xnnsLet:n=2k=2∑k=1∞x2k(2k)s=21−s∑k=1∞(x2)kks=21−sLis(x2)\displaystyle \begin{aligned} \mathrm{Li}_s(x)+\mathrm{Li}_s(-x)&=\sum_{n=1}^{\infty}\frac{x^n}{n^s}+\sum_{n=1}^{\infty}\frac{(-x)^n}{n^s} \\ &=2\sum_{n=2,4,\cdots}^{\infty}\frac{x^n}{n^s}\qquad\quad\mathrm{Let}: n=2k \\ &=2\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k)^s} \\ &=2^{1-s}\sum_{k=1}^{\infty}\frac{(x^2)^k}{k^s} \\ &=2^{1-s}\mathrm{Li}_s(x^2) \end{aligned}
  • Euler's Reflection Formula

    Li2(x)+Li2(1−x)=π26−ln⁡xln⁡(1−x)\displaystyle \mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}6-\ln x\ln{(1-x)}
    [증명]

    Li2(x)+Li2(1−x)=Li2(x)+∫01−x−ln⁡(1−t)t dtLet:t=1−u=Li2(x)−∫1xln⁡u⋅−du1−u=Li2(x)−ln⁡uln⁡(1−u)∣1x+∫1x1uln⁡(1−u) du=∫0x−ln⁡(1−u)u du+∫x1−ln⁡(1−u)u du−ln⁡xln⁡(1−x)+lim⁡u→1ln⁡uln⁡(1−u)=∫01−ln⁡(1−u)u du−ln⁡xln⁡(1−x)+0=Li2(1)−ln⁡xln⁡(1−x)=ζ(2)−ln⁡xln⁡(1−x)=π26−ln⁡xln⁡(1−x)\displaystyle \begin{aligned} \mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)&=\mathrm{Li}_2(x)+\int_0^{1-x}\frac{-\ln{(1-t)}}t\,\mathrm{d}t\qquad\quad\mathrm{Let}: t=1-u \\ &=\mathrm{Li}_2(x)-\int_1^x\ln u\cdot\frac{-\mathrm{d}u}{1-u} \\ &=\mathrm{Li}_2(x)-\bigl.\ln u\ln{(1-u)}\bigr|_1^x+\int_1^x\frac1u\ln{(1-u)}\,\mathrm{d}u \\ &=\int_0^x\frac{-\ln{(1-u)}}u\,\mathrm{d}u+\int_x^1\frac{-\ln{(1-u)}}u\,\mathrm{d}u-\ln x\ln{(1-x)}+\lim_{u\to1}\ln u\ln{(1-u)} \\ &=\int_0^1\frac{-\ln{(1-u)}}u\,\mathrm{d}u-\ln x\ln{(1-x)}+0 \\ &=\mathrm{Li}_2(1)-\ln x\ln{(1-x)} \\ &=\zeta(2)-\ln x\ln{(1-x)} \\ &=\frac{\pi^2}6-\ln x\ln{(1-x)} \end{aligned}

    위 증명 과정의 넷째 줄에 있는 극한값은 아래와 같이 계산되었다. 로피탈의 정리를 사용한 곳은 =∗\overset{*}=로 나타내었다.
    먼저 1−u=t1-u=t로 치환함으로써 시작한다.

    lim⁡u→1ln⁡uln⁡(1−u)=lim⁡t→0ln⁡(1−t)ln⁡t=lim⁡t→0ln⁡t1/tln⁡(1−t)t=lim⁡t→0ln⁡t1/tlim⁡t→0ln⁡(1−t)t=∗lim⁡t→01/t−1/t2lim⁡t→0−1/(1−t)1=lim⁡t→0tlim⁡t→011−t=0⋅1=0\displaystyle \begin{aligned} \lim_{u\to1}\ln u\ln{(1-u)}&=\lim_{t\to0}\ln{(1-t)}\ln t \\ &=\lim_{t\to0}\frac{\ln t}{1/t}\frac{\ln{(1-t)}}{t} \\ &=\lim_{t\to0}\frac{\ln t}{1/t}\lim_{t\to0}\frac{\ln{(1-t)}}{t} \\ &\overset{*}=\lim_{t\to0}\frac{1/t}{-1/t^2}\lim_{t\to0}\frac{-1/(1-t)}{1} \\ &=\lim_{t\to0}t\lim_{t\to0}\frac1{1-t} \\ &=0\cdot1 \\ &=0 \end{aligned}
  • Inversion Formula

    Li2(x)+Li2 ⁣(1x)=−π26−12ln⁡2(−x)\displaystyle \mathrm{Li}_2(x)+\mathrm{Li}_2\!\left(\frac1x\right)=-\frac{\pi^2}6-\frac12\ln^2(-x)
    [증명]

    Li2 ⁣(−1x)=∫0−1x−ln⁡(1−t)t dt⇒ddxLi2 ⁣(−1x)=−ln⁡ ⁣(1+1x)−1x⋅1x2=ln⁡(x+1)−ln⁡xx⇒∫1tddxLi2 ⁣(−1x)dx=Li2 ⁣(−1t)−Li2(−1)=∫1tln⁡(x+1)−ln⁡xx dx=∫1t−ln⁡(1−(−x))−x dx−[12ln⁡2x]1tLet:−x=y=∫−1−t−ln⁡(1−y)y(−dy)−12ln⁡2t=∫−t−1−ln⁡(1−y)y dy−12ln⁡2t=∫0−1−ln⁡(1−y)y dy−∫0−t−ln⁡(1−y)y dy−12ln⁡2t=Li2(−1)−Li2(−t)−12ln⁡2t∴Li2 ⁣(−1t)−Li2(−1)=Li2(−1)−Li2(−t)−12ln⁡2t⇒Li2(−t)+Li2 ⁣(−1t)=2Li2(−1)−12ln⁡2t=2⋅(2−1−1)ζ(2)−12ln⁡2t=−π26−12ln⁡2tLet:t=−x∴Li2(x)+Li2 ⁣(1x)=−π26−12ln⁡2(−x)\displaystyle \begin{aligned} \mathrm{Li}_2\!\left(-\frac1x\right)&=\int_0^{-\frac1x}\frac{-\ln{(1-t)}}t\,\mathrm{d}t \\ \Rightarrow\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_2\!\left(-\frac1x\right)&=\frac{-\ln\!\left(1+\frac1x\right)}{-\frac1x}\cdot\frac1{x^2} \\ &=\frac{\ln{(x+1)}-\ln x}x \\ \Rightarrow\int_1^t\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_2\!\left(-\frac1x\right)\mathrm{d}x&=\mathrm{Li}_2\!\left(-\frac1t\right)-\mathrm{Li}_2(-1) \\ &=\int_1^t\frac{\ln{(x+1)}-\ln x}x\,\mathrm{d}x \\ &=\int_1^t\frac{-\ln{(1-(-x))}}{-x}\,\mathrm{d}x-\biggl[\frac12\ln^2x\biggr]_1^t \\ &\qquad\quad\mathrm{Let}: -x=y \\ &=\int_{-1}^{-t}\frac{-\ln{(1-y)}}y(-\mathrm{d}y)-\frac12\ln^2t \\ &=\int_{-t}^{-1}\frac{-\ln{(1-y)}}y\,\mathrm{d}y-\frac12\ln^2t \\ &=\int_{0}^{-1}\frac{-\ln{(1-y)}}y\,\mathrm{d}y-\int_{0}^{-t}\frac{-\ln{(1-y)}}y\,\mathrm{d}y-\frac12\ln^2t \\ &=\mathrm{Li}_2(-1)-\mathrm{Li}_2(-t)-\frac12\ln^2t \\ \therefore\mathrm{Li}_2\!\left(-\frac1t\right)-\mathrm{Li}_2(-1)&=\mathrm{Li}_2(-1)-\mathrm{Li}_2(-t)-\frac12\ln^2t \\ \Rightarrow\mathrm{Li}_2(-t)+\mathrm{Li}_2\!\left(-\frac1t\right)&=2\mathrm{Li}_2(-1)-\frac12\ln^2t \\ &=2\cdot(2^{-1}-1)\zeta(2)-\frac12\ln^2t \\ &=-\frac{\pi^2}6-\frac12\ln^2t\qquad\quad\mathrm{Let}: t=-x \\ \therefore\mathrm{Li}_2(x)+\mathrm{Li}_2\!\left(\frac1x\right)&=-\frac{\pi^2}6-\frac12\ln^2(-x) \end{aligned}
  • Landen's Identity

    Li2(1−x)+Li2 ⁣(1−1x)=−12ln⁡2x\displaystyle \mathrm{Li}_2(1-x)+\mathrm{Li}_2\!\left(1-\frac1x\right)=-\frac12\ln^2x
    [증명]

    Li2 ⁣(x1+x)=∫0x1+x−ln⁡(1−t)t dt⇒ddxLi2 ⁣(x1+x)=−ln⁡ ⁣(1−x1+x)x1+x⋅1(1+x)2=−ln⁡ ⁣(11+x)x(1+x)=ln⁡(1+x)⋅(1x−11+x)⇒∫0tddxLi2 ⁣(x1+x)dx=Li2 ⁣(t1+t)=∫0tln⁡(1+x)x dx−∫0tln⁡(1+x)1+x dx=∫0t−ln⁡(1−(−x))−x dx−[12ln⁡2(1+x)]0tLet:−x=y=∫0−t−ln⁡(1−y)y(−dy)−12ln⁡2(1+t)=−Li2(−t)−12ln⁡2(1+t)∴Li2 ⁣(t1+t)=−Li2(−t)−12ln⁡2(1+t)Let:1+t=x⇒Li2 ⁣(1−1x)=−Li2(1−x)−12ln⁡2x∴Li2(1−x)+Li2 ⁣(1−1x)=−12ln⁡2x\displaystyle \begin{aligned} \mathrm{Li}_2\!\left(\frac x{1+x}\right)&=\int_0^{\textstyle \frac x{1+x}}\frac{-\ln{(1-t)}}t\,\mathrm{d}t \\ \Rightarrow\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_2\!\left(\frac x{1+x}\right)&=\frac{-\ln\!\left(1-\frac x{1+x}\right)}{\frac x{1+x}}\cdot\frac1{(1+x)^2} \\ &=-\frac{\ln\!\left(\frac1{1+x}\right)}{x(1+x)} \\ &=\ln{(1+x)}\cdot\left(\frac1x-\frac1{1+x}\right) \\ \Rightarrow\int_0^t\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_2\!\left(\frac x{1+x}\right)\mathrm{d}x&=\mathrm{Li}_2\!\left(\frac t{1+t}\right) \\ &=\int_0^t\frac{\ln{(1+x)}}x\,\mathrm{d}x-\int_0^t\frac{\ln{(1+x)}}{1+x}\,\mathrm{d}x \\ &=\int_0^t\frac{-\ln{(1-(-x))}}{-x}\,\mathrm{d}x-\biggl[\frac12\ln^2(1+x)\biggr]_0^t \\ &\qquad\quad\mathrm{Let}: -x=y \\ &=\int_0^{-t}\frac{-\ln{(1-y)}}y(-\mathrm{d}y)-\frac12\ln^2(1+t) \\ &=-\mathrm{Li}_2(-t)-\frac12\ln^2(1+t) \\ \therefore\mathrm{Li}_2\!\left(\frac t{1+t}\right)&=-\mathrm{Li}_2(-t)-\frac12\ln^2(1+t)\qquad\quad\mathrm{Let}: 1+t=x \\ \Rightarrow\mathrm{Li}_2\!\left(1-\frac1x\right)&=-\mathrm{Li}_2(1-x)-\frac12\ln^2x \\ \therefore\mathrm{Li}_2(1-x)+\mathrm{Li}_2\!\left(1-\frac1x\right)&=-\frac12\ln^2x \end{aligned}
  • Inversion Formula for Trilogarithm

    Li3(x)−Li3 ⁣(1x)=−π26ln⁡(−x)−16ln⁡3(−x)\displaystyle \mathrm{Li}_3(x)-\mathrm{Li}_3\!\left(\frac1x\right)=-\frac{\pi^2}6\ln{(-x)}-\frac16\ln^3(-x)
    [증명]

    Li3 ⁣(−1x)=∫0−1xLi2(t)t dt⇒ddxLi3 ⁣(−1x)=Li2 ⁣(−1x)−1x⋅1x2=−Li2 ⁣(−1x)x(Use Inversion Formula)=−1x[−Li2(−x)−π26−12ln⁡2x]=Li2(−x)x+π261x+12ln⁡2xx⇒∫1tddxLi3 ⁣(−1x)dx=Li3 ⁣(−1t)−Li3(−1)=∫1tLi2(−x)x dx+π26∫1tdxx+12∫1tln⁡2xx dxLet:−x=y⇒dxx=dyy=∫−1−tLi2(y)y dy+π26[ln⁡x]1t+12[13ln⁡3x]1t=∫0−tLi2(y)y dy−∫0−1Li2(y)y dy+π26ln⁡t+16ln⁡3t=Li3(−t)−Li3(−1)+π26ln⁡t+16ln⁡3t∴Li3 ⁣(−1t)−Li3(−1)=Li3(−t)−Li3(−1)+π26ln⁡t+16ln⁡3t⇒Li3 ⁣(−1t)−Li3(−t)=π26ln⁡t+16ln⁡3tLet:t=−x∴Li3(x)−Li3 ⁣(1x)=−π26ln⁡(−x)−16ln⁡3(−x)\displaystyle \begin{aligned} \mathrm{Li}_3\!\left(-\frac1x\right)&=\int_0^{-\frac1x}\frac{\mathrm{Li}_2(t)}t\,\mathrm{d}t \\ \Rightarrow\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_3\!\left(-\frac1x\right)&=\frac{\mathrm{Li}_2\!\left(-\frac1x\right)}{-\frac1x}\cdot\frac1{x^2} \\ &=-\frac{\mathrm{Li}_2\!\left(-\frac1x\right)}x\qquad\quad(\mathrm{Use\ Inversion\ Formula}) \\ &=-\frac1x\left[-\mathrm{Li}_2(-x)-\frac{\pi^2}6-\frac12\ln^2x\right] \\ &=\frac{\mathrm{Li}_2(-x)}x+\frac{\pi^2}6\frac1x+\frac12\frac{\ln^2x}x \\ \Rightarrow\int_1^t\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_3\!\left(-\frac1x\right)\mathrm{d}x&=\mathrm{Li}_3\!\left(-\frac1t\right)-\mathrm{Li}_3(-1) \\ &=\int_1^t\frac{\mathrm{Li}_2(-x)}x\,\mathrm{d}x+\frac{\pi^2}6\int_1^t\frac{\mathrm{d}x}x+\frac12\int_1^t\frac{\ln^2x}x\,\mathrm{d}x \\ &\qquad\quad\mathrm{Let}: -x=y\Rightarrow\frac{\mathrm{d}x}x=\frac{\mathrm{d}y}y \\ &=\int_{-1}^{-t}\frac{\mathrm{Li}_2(y)}y\,\mathrm{d}y+\frac{\pi^2}6\bigl[\ln x\bigr]_1^t+\frac12\biggl[\frac13\ln^3x\biggr]_1^t \\ &=\int_{0}^{-t}\frac{\mathrm{Li}_2(y)}y\,\mathrm{d}y-\int_{0}^{-1}\frac{\mathrm{Li}_2(y)}y\,\mathrm{d}y+\frac{\pi^2}6\ln t+\frac16\ln^3t \\ &=\mathrm{Li}_3(-t)-\mathrm{Li}_3(-1)+\frac{\pi^2}6\ln t+\frac16\ln^3t \\ \therefore\mathrm{Li}_3\!\left(-\frac1t\right)-\mathrm{Li}_3(-1)&=\mathrm{Li}_3(-t)-\mathrm{Li}_3(-1)+\frac{\pi^2}6\ln t+\frac16\ln^3t \\ \Rightarrow\mathrm{Li}_3\!\left(-\frac1t\right)-\mathrm{Li}_3(-t)&=\frac{\pi^2}6\ln t+\frac16\ln^3t\qquad\quad\mathrm{Let}: t=-x \\ \therefore\mathrm{Li}_3(x)-\mathrm{Li}_3\!\left(\frac1x\right)&=-\frac{\pi^2}6\ln{(-x)}-\frac16\ln^3(-x) \end{aligned}

4. 알려진 함숫값[편집]

  • Li1 ⁣(12)=∑n=1∞12nn=ln⁡2≈0.6931471806\displaystyle \mathrm{Li}_1\!\left(\frac12\right)=\sum_{n=1}^{\infty}\frac1{2^nn}=\ln2\approx0.6931471806
    [증명]

    위의 Li1(x)\mathrm{Li}_1(x)에 관한 식에 x=1/2x=1/2을 대입하면 된다.

    Li1 ⁣(12)=−ln⁡ ⁣(1−12)=ln⁡2\displaystyle \mathrm{Li}_1\!\left(\dfrac12\right)=-\ln\!\left(1-\dfrac12\right)=\ln2

  • Li2 ⁣(12)=∑n=1∞12nn2=−12ln⁡22+π212≈0.5822405265\displaystyle \mathrm{Li}_2\!\left(\frac12\right)=\sum_{n=1}^{\infty}\frac1{2^nn^2}=-\frac12\ln^22+\frac{\pi^2}{12}\approx0.5822405265
    [증명]

    로그 함수의 매클로린 급수에서 시작한다.

    ln⁡(1−x)=−∑n=1∞xnnln⁡(1−x)x=−∑n=1∞xn−1n∫01/2ln⁡(1−x)x dx=−∫01/2∑n=1∞xn−1n dx=−∑n=1∞∫01/2xn−1n dx=−∑n=1∞xnn2∣01/2=−∑n=1∞12nn2=−Li2 ⁣(12)⋯(1)\displaystyle \begin{aligned} \ln{(1-x)}&=-\sum_{n=1}^{\infty}\frac{x^n}n \\ \frac{\ln{(1-x)}}x&=-\sum_{n=1}^{\infty}\frac{x^{n-1}}n \\ \int_0^{1/2}\frac{\ln{(1-x)}}x\,\mathrm{d}x&=-\int_0^{1/2}\sum_{n=1}^{\infty}\frac{x^{n-1}}n\,\mathrm{d}x=-\sum_{n=1}^{\infty}\int_0^{1/2}\frac{x^{n-1}}n\,\mathrm{d}x \\ &=-\sum_{n=1}^{\infty}\left.\frac{x^n}{n^2}\right|_0^{1/2}=-\sum_{n=1}^{\infty}\frac1{2^n n^2} \\ &=-\mathrm{Li}_2\!\left(\frac12\right)\quad\cdots(1) \end{aligned}

    한편, 좌변의 정적분은 다음과 같이 계산할 수 있다. 부분적분으로 시작한다.

    ∫01/2ln⁡(1−x)x dx=∫01/2ln⁡(1−x)⋅1x dx=ln⁡(1−x)ln⁡x∣01/2+∫01/2ln⁡x1−x dxLet:x=1−t=(ln⁡212−lim⁡x→0+ln⁡(1−x)ln⁡x)−∫11/2ln⁡(1−t)t dt=ln⁡22−0−∫11/2ln⁡(1−x)x dx=ln⁡22+∑n=1∞xnn2∣11/2=ln⁡22+∑n=1∞12nn2−∑n=1∞1n2=ln⁡22+Li2 ⁣(12)−ζ(2)=ln⁡22+Li2 ⁣(12)−π26⋯(2)\displaystyle \begin{aligned} \int_0^{1/2}\frac{\ln{(1-x)}}x\,\mathrm{d}x&=\int_0^{1/2}\ln{(1-x)}\cdot\frac1x\,\mathrm{d}x \\ &=\Bigl.\ln{(1-x)}\ln x\Bigr|_0^{1/2}+\int_0^{1/2}\frac{\ln x}{1-x}\,\mathrm{d}x\qquad\quad\mathrm{Let}: x=1-t \\ &=\left(\ln^2\frac12-\lim_{x\to0^+}\ln{(1-x)}\ln x\right)-\int_1^{1/2}\frac{\ln{(1-t)}}{t}\,\mathrm{d}t \\ &=\ln^22-0-\int_1^{1/2}\frac{\ln{(1-x)}}x\,\mathrm{d}x=\ln^22+\sum_{n=1}^{\infty}\left.\frac{x^n}{n^2}\right|_1^{1/2} \\ &=\ln^22+\sum_{n=1}^{\infty}\frac1{2^nn^2}-\sum_{n=1}^{\infty}\frac1{n^2}=\ln^22+\mathrm{Li}_2\!\left(\frac12\right)-\zeta(2) \\ &=\ln^22+\mathrm{Li}_2\!\left(\frac12\right)-\frac{\pi^2}6\quad\cdots(2) \end{aligned}

    위 식의 셋째 줄에 있는 극한값은 아래와 같이 계산되었다. 로피탈의 정리를 사용한 곳은 \overset\mathbf{*}=로 나타내었다.

    \displaystyle \begin{aligned} \lim_{x\to0^+}\ln{(1-x)}\ln x&=\lim_{x\to0^+}\frac{\ln x}{1/x}\frac{\ln{(1-x)}}{x} \\& =\lim_{x\to0^+}\frac{\ln x}{1/x}\lim_{x\to0^+}\frac{\ln{(1-x)}}{x} \\ &\overset\mathbf*=\lim_{x\to0^+}\frac{1/x}{-1/x^{2}}\lim_{x\to0^+}\frac{-1/(1-x)}{1}\\&=\lim_{x\to0^+}x\lim_{x\to0^+}\frac1{1-x} \\&=0\cdot1 \\&=0 \end{aligned}

    (1)(1)과 (2)(2)에 따라 다음과 같이 Li2(1/2)\mathrm{Li}_2(1/2)의 값을 계산할 수 있다.

    ln⁡22+Li2 ⁣(12)−π26=−Li2 ⁣(12)2Li2 ⁣(12)=π26−ln⁡22∴Li2 ⁣(12)=−12ln⁡22+π212\displaystyle \begin{aligned} \ln^22+\mathrm{Li}_2\!\left(\frac12\right)-\frac{\pi^2}6&=-\mathrm{Li}_2\!\left(\frac12\right) \\ 2\mathrm{Li}_2\!\left(\frac12\right)&=\frac{\pi^2}6-\ln^22 \\ \therefore\mathrm{Li}_2\!\left(\frac12\right)&=-\frac12\ln^22+\frac{\pi^2}{12} \end{aligned}

  • Li3 ⁣(12)=∑n=1∞12nn3=16ln⁡32−π212ln⁡2+78 ζ(3)≈0.5372131936\displaystyle \mathrm{Li}_3\!\left(\frac12\right)=\sum_{n=1}^{\infty}\frac1{2^nn^3}=\frac16 \ln^3 2-\frac{\pi^2}{12}\ln2+\frac78\,\zeta(3)\approx0.5372131936
    [증명]

    먼저 다음 4가지의 참고식들을 증명한 다음, 이 참고식들을 사용하여 Li3 ⁣(12)\mathrm{Li}_3\!\left(\dfrac12\right)의 값을 계산할 것이다.

    • 참고식 AA, BB
      n∈Z+n\in\mathbb{Z}^+, k∈Z+k\in\mathbb{Z}^+에 대하여 다음 두 식을 얻을 수 있다. 여기서 Z+\mathbb{Z}^+는 양의 정수의 집합을 의미한다.

      ∫01xn−1ln⁡kx dx=1nxnln⁡kx∣01−∫011nxn⋅kln⁡k−1x⋅1x dx=−kn∫01xn−1ln⁡k−1x dx=⋯=(−1)kk!nk∫01xn−1 dx=(−1)kk!nk+1⋯A⇒1nk+1=(−1)kk!∫01xn−1ln⁡kx dx⋯B\displaystyle \begin{aligned} \int_0^1x^{n-1}\ln^kx\,\mathrm{d}x&=\Bigl.\frac1nx^n\ln^kx\Bigr|_0^1-\int_0^1\frac1nx^n\cdot k\ln^{k-1}x\cdot\frac1x\,\mathrm{d}x \\ &=-\frac kn\int_0^1x^{n-1}\ln^{k-1}x\,\mathrm{d}x \\ &=\cdots \\ &=\frac{(-1)^kk!}{n^k}\int_0^1x^{n-1}\,\mathrm{d}x \\ &=\frac{(-1)^kk!}{n^{k+1}}\qquad\cdots A \\ \Rightarrow\frac1{n^{k+1}}&=\frac{(-1)^k}{k!}\int_0^1x^{n-1}\ln^kx\,\mathrm{d}x\qquad\cdots B \end{aligned}


    • 참고식 CC
      s∈Z+s\in\mathbb{Z}^+에 대하여 다음 식을 얻을 수 있다.

      ∑n=1∞(−1)n−1ns=∑n=1∞(−1)n+1ns=(1−21−s)ζ(s)⋯C\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^s}=(1-2^{1-s})\zeta(s)\qquad\cdots C

      증명은 제타 함수 문서의 성질 문단에서 볼 수 있다.

    • 참고식 DD
      먼저 x=t2x=t^2으로 치환함으로써 시작한다.

      ∫01ln⁡2x1−x dx=∫01(2ln⁡t)21−t2 2t dt=8∫01ln⁡2t⋅t1−t2 dt=8∫01ln⁡2t⋅12(11−t−11+t)dt=4∫01ln⁡2t1−t dt−4∫01ln⁡2t1+t dt⇒−3∫01ln⁡2x1−x dx=−4∫01ln⁡2t1+t dt∴∫01ln⁡2x1−x dx=43∫01ln⁡2t1+t dt⋯D\displaystyle \begin{aligned} \int_0^1\frac{\ln^2x}{1-x}\,\mathrm{d}x&=\int_0^1\frac{(2\ln t)^2}{1-t^2}\,2t\,\mathrm{d}t \\ &=8\int_0^1\ln^2t\cdot\frac t{1-t^2}\,\mathrm{d}t \\ &=8\int_0^1\ln^2t\cdot\frac12\left(\frac1{1-t}-\frac1{1+t}\right)\mathrm{d}t \\ &=4\int_0^1\frac{\ln^2t}{1-t}\,\mathrm{d}t-4\int_0^1\frac{\ln^2t}{1+t}\,\mathrm{d}t \\ \Rightarrow-3\int_0^1\frac{\ln^2x}{1-x}\,\mathrm{d}x&=-4\int_0^1\frac{\ln^2t}{1+t}\,\mathrm{d}t \\ \therefore\int_0^1\frac{\ln^2x}{1-x}\,\mathrm{d}x&=\frac43\int_0^1\frac{\ln^2t}{1+t}\,\mathrm{d}t\qquad\cdots D \end{aligned}



    이제 위의 참고식들을 사용하여 Li3 ⁣(12)\mathrm{Li}_3\!\left(\dfrac12\right)의 값을 계산해보자. 계산 과정 중, 위의 참고식이나 아래의 과정에서 나타나는 관계식을 사용한 경우, 등호 위에다가 해당 참고식·관계식의 알파벳·번호를 적을 것이다. (예: =A\overset{A}=, =(1)\overset{(1)}=)

    Li3 ⁣(12)=∑n=1∞12nn3=B∑n=1∞12n(12∫01xn−1ln⁡2x dx)=14∑n=1∞12n−1∫01xn−1ln⁡2x dx=14∫01∑n=1∞(x2)n−1ln⁡2x dx=14∫0111−x2ln⁡2x dx=12∫01ln⁡2x2−x dxLet:x2−x=t→x=2t1+t→dx=2(1+t)2 dt=12∫01ln⁡2 ⁣(2t1+t)2−(2t1+t)2(1+t)2 dt=12∫01ln⁡2 ⁣(2t1+t)1+t dt=12∫01(ln⁡2+ln⁡t−ln⁡(1+t))21+t dt=12∫01ln⁡22+ln⁡2t+ln⁡2(1+t)+2ln⁡2ln⁡t−2ln⁡tln⁡(1+t)−2ln⁡2ln⁡(1+t)1+t dt=12∫01ln⁡22−2ln⁡2ln⁡(1+t)+ln⁡2t+ln⁡2(1+t)+2ln⁡2ln⁡t−2ln⁡tln⁡(1+t)1+t dt⋯(1)\displaystyle \begin{aligned} \mathrm{Li}_3\!\left(\frac12\right)&=\sum_{n=1}^{\infty}\frac1{2^n n^3}\overset{B}=\sum_{n=1}^{\infty}\frac1{2^n}\left(\frac12\int_0^1x^{n-1}\ln^2x\,\mathrm{d}x\right)=\frac14\sum_{n=1}^{\infty}\frac1{2^{n-1}}\int_0^1x^{n-1}\ln^2x\,\mathrm{d}x \\ &=\frac14\int_0^1\sum_{n=1}^{\infty}\left(\frac x2\right)^{n-1}\ln^2x\,\mathrm{d}x=\frac14\int_0^1\frac1{1-\frac x2}\ln^2x\,\mathrm{d}x=\frac12\int_0^1\frac{\ln^2x}{2-x}\,\mathrm{d}x \\ &\qquad\quad\mathrm{Let}:\frac x{2-x}=t\quad\rightarrow\quad x=\frac{2t}{1+t}\quad\rightarrow\quad\mathrm{d}x=\frac2{(1+t)^2}\,\mathrm{d}t \\ &=\frac12\int_0^1\frac{\ln^2\!\left(\dfrac{2t}{1+t}\right)}{2-\left(\dfrac{2t}{1+t}\right)}\frac2{(1+t)^2}\,\mathrm{d}t=\frac12\int_0^1\frac{\ln^2\!\left(\dfrac{2t}{1+t}\right)}{1+t}\,\mathrm{d}t \\ &=\frac12\int_0^1\frac{(\ln2+\ln t-\ln{(1+t)})^2}{1+t}\,\mathrm{d}t \\ &=\frac12\int_0^1\frac{\ln^22+\ln^2t+\ln^2(1+t)+2\ln2\ln t-2\ln t\ln{(1+t)}-2\ln2\ln{(1+t)}}{1+t}\,\mathrm{d}t \\ &=\frac12\int_0^1\frac{{\color{blue}\ln^22-2\ln2\ln{(1+t)}}+{\color{red}\ln^2t}+{\color{limegreen}\ln^2(1+t)}+{\color{darkorchid}2\ln2\ln t}-{\color{fuchsia}2\ln t\ln{(1+t)} }}{1+t}\,\mathrm{d}t\qquad\cdots(1) \end{aligned}


    위의 정적분을 색깔별로 쪼개서 적분한 후 마지막에 값을 합쳐주면 된다.

    ∫01ln⁡22−2ln⁡2ln⁡(1+t)1+t dt=ln⁡2∫01ln⁡2−2ln⁡(1+t)1+t dt=letILet:t=1−x1+x→1+t=21+x→dt=−2(1+x)2 dx=ln⁡2∫10ln⁡2−2ln⁡ ⁣(21+x)21+x−2(1+x)2 dx=ln⁡2∫01ln⁡2−2ln⁡2+2ln⁡(1+x)1+x dx=ln⁡2∫01−ln⁡2+2ln⁡(1+x)1+x dx=−I⇒I=−I∴I=0=0⋯(2)\displaystyle \begin{aligned} \int_0^1\frac{{\color{blue}\ln^22-2\ln2\ln{(1+t)} }}{1+t}\,\mathrm{d}t&=\ln2\int_0^1\frac{\ln2-2\ln{(1+t)}}{1+t}\,\mathrm{d}t\overset{\mathrm{let}}=I \\ &\qquad\mathrm{Let}:t=\frac{1-x}{1+x}\quad\rightarrow\quad1+t=\frac2{1+x}\quad\rightarrow\quad\mathrm{d}t=\frac{-2}{(1+x)^2}\,\mathrm{d}x \\ &=\ln2\int_1^0\frac{\ln2-2\ln\!\left(\dfrac2{1+x}\right)}{\cfrac2{1+x}}\frac{-2}{(1+x)^2}\,\mathrm{d}x \\ &=\ln2\int_0^1\frac{\ln2-2\ln2+2\ln{(1+x)}}{1+x}\,\mathrm{d}x \\ &=\ln2\int_0^1\frac{-\ln2+2\ln{(1+x)}}{1+x}\,\mathrm{d}x=-I\quad\Rightarrow I=-I\quad\therefore I=0 \\ &=0\qquad\cdots(2) \end{aligned}


    ∫01ln⁡2t1+t dt=∫01ln⁡2t∑n=1∞(−1)n−1 tn−1 dt=∑n=1∞(−1)n−1∫01tn−1ln⁡2t dt=A∑n=1∞(−1)n−1(2n3)=2∑n=1∞(−1)n−1n3=C2⋅34ζ(3)=32ζ(3)⋯(3)\displaystyle \begin{aligned} \int_0^1\frac{{\color{red}\ln^2t}}{1+t}\,\mathrm{d}t&=\int_0^1\ln^2t\sum_{n=1}^{\infty}(-1)^{n-1}\,t^{n-1}\,\mathrm{d}t \\ &=\sum_{n=1}^{\infty}(-1)^{n-1}\int_0^1t^{n-1}\ln^2t\,\mathrm{d}t \\ &\overset{A}=\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac2{n^3}\right) \\ &=2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3} \\ &\overset{C}=2\cdot\frac34\zeta(3) \\ &=\frac32\zeta(3)\qquad\cdots(3) \end{aligned}


    ∫01ln⁡2(1+t)1+t dt=13ln⁡3(1+t)∣01=13ln⁡32⋯(4)\displaystyle \begin{aligned} \int_0^1\frac{{\color{limegreen}\ln^2(1+t)}}{1+t}\,\mathrm{d}t&=\Bigl.\frac13\ln^3(1+t)\Bigr|_0^1 \\ &=\frac13\ln^32\qquad\cdots(4) \end{aligned}


    ∫012ln⁡2ln⁡t1+t dt=2ln⁡2∫01ln⁡t∑n=1∞(−1)n−1 tn−1 dt=2ln⁡2∑n=1∞(−1)n−1∫01tn−1ln⁡t dt=A2ln⁡2∑n=1∞(−1)n−1(−1n2)=−2ln⁡2∑n=1∞(−1)n−1n2=C−2ln⁡2⋅12ζ(2)=−π26ln⁡2⋯(5)\displaystyle \begin{aligned} \int_0^1\frac{{\color{darkorchid}2\ln2\ln t}}{1+t}\,\mathrm{d}t&=2\ln2\int_0^1\ln t\sum_{n=1}^{\infty}(-1)^{n-1}\,t^{n-1}\,\mathrm{d}t \\ &=2\ln2\sum_{n=1}^{\infty}(-1)^{n-1}\int_0^1t^{n-1}\ln t\,\mathrm{d}t \\ &\overset{A}=2\ln2\sum_{n=1}^{\infty}(-1)^{n-1}\left(-\frac1{n^2}\right) \\ &=-2\ln2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} \\ &\overset{C}=-2\ln2\cdot\frac12\zeta(2) \\ &=-\frac{\pi^2}6\ln2\qquad\cdots(5) \end{aligned}


    ∫012ln⁡tln⁡(1+t)1+t dt=∫01ln⁡t⋅2ln⁡(1+t)1+t dt=ln⁡tln⁡2(1+t)∣01−∫011t⋅ln⁡2(1+t) dt=−∫01ln⁡2(1+t)t dt=−12∫01(ln⁡(1−t)+ln⁡(1+t))2t dt−12∫01(ln⁡(1−t)−ln⁡(1+t))2t dt+∫01ln⁡2(1−t)t dt=−12∫01ln⁡2(1−t2)t dt−12∫01ln⁡2 ⁣(1−t1+t)t dt+∫01ln⁡2(1−t)t dtLet:1−t2=x→t2=1−x→2t dt=−dx→dtt=−12dxt2=−12dx1−xLet:1−t1+t=y→t=1−y1+y→dt=−2(1+y)2 dy→dtt=1+y1−y−2(1+y)2dy=−(11−y+11+y)dyLet:1−t=z→t=1−z→dt=−dz→dtt=−dz1−z=14∫10ln⁡2x1−x dx+12∫10(ln⁡2y1−y+ln⁡2y1+y)dy−∫10ln⁡2z1−z dz=14∫01ln⁡2x1−x dx−12∫01ln⁡2y1+y dy=D14⋅43∫01ln⁡2x1+x dx−12∫01ln⁡2y1+y dy=−16∫01ln⁡2x1+x dx=(3)−16⋅32ζ(3)=−14ζ(3)⋯(6)\displaystyle \begin{aligned} \int_0^1\frac{{\color{fuchsia}2\ln t\ln{(1+t)} }}{1+t}\,\mathrm{d}t&=\int_0^1\ln t\cdot\frac{2\ln{(1+t)}}{1+t}\,\mathrm{d}t \\ &=\Bigl.\ln t\ln^2(1+t)\Bigr|_0^1-\int_0^1\frac1t\cdot\ln^2(1+t)\,\mathrm{d}t \\ &=-\int_0^1\frac{\ln^2(1+t)}t\,\mathrm{d}t \\ &=-\frac12\int_0^1\frac{(\ln{(1-t)}+\ln{(1+t)})^2}t\,\mathrm{d}t-\frac12\int_0^1\frac{(\ln{(1-t)}-\ln{(1+t)})^2}t\,\mathrm{d}t+\int_0^1\frac{\ln^2(1-t)}t\,\mathrm{d}t \\ &=-\frac12\int_0^1\frac{\ln^2(1-t^2)}t\,\mathrm{d}t-\frac12\int_0^1\frac{\ln^2\!\left(\dfrac{1-t}{1+t}\right)}t\,\mathrm{d}t+\int_0^1\frac{\ln^2(1-t)}{t}\,\mathrm{d}t \\ &\qquad\quad\mathrm{Let}:1-t^2=x\quad\rightarrow\quad t^2=1-x\quad\rightarrow\quad2t\,\mathrm{d}t=-\mathrm{d}x \\ &\qquad\qquad\rightarrow\quad\frac{\mathrm{d}t}t=-\frac12\frac{\mathrm{d}x}{t^2}=-\frac12\frac{\mathrm{d}x}{1-x} \\ &\qquad\quad\mathrm{Let}:\frac{1-t}{1+t}=y\quad\rightarrow\quad t=\frac{1-y}{1+y}\quad\rightarrow\quad\mathrm{d}t=\frac{-2}{(1+y)^2}\,\mathrm{d}y \\ &\qquad\qquad\rightarrow\quad\frac{\mathrm{d}t}t=\frac{1+y}{1-y}\frac{-2}{(1+y)^2}\mathrm{d}y=-\left(\frac1{1-y}+\frac1{1+y}\right)\mathrm{d}y \\ &\qquad\quad\mathrm{Let}:1-t=z\quad\rightarrow\quad t=1-z\quad\rightarrow\quad\mathrm{d}t=-\mathrm{d}z \\ &\qquad\qquad\rightarrow\quad\frac{\mathrm{d}t}t=-\frac{\mathrm{d}z}{1-z} \\ &=\frac14\int_1^0\frac{\ln^2x}{1-x}\,\mathrm{d}x+\frac12\int_1^0\left(\frac{\ln^2y}{1-y}+\frac{\ln^2y}{1+y}\right)\mathrm{d}y-\int_1^0\frac{\ln^2z}{1-z}\,\mathrm{d}z \\ &=\frac14\int_0^1\frac{\ln^2x}{1-x}\,\mathrm{d}x-\frac12\int_0^1\frac{\ln^2y}{1+y}\,\mathrm{d}y \\ &\overset{D}=\frac14\cdot\frac43\int_0^1\frac{\ln^2x}{1+x}\,\mathrm{d}x-\frac12\int_0^1\frac{\ln^2y}{1+y}\,\mathrm{d}y \\ &=-\frac16\int_0^1\frac{\ln^2x}{1+x}\,\mathrm{d}x \\ &\overset{(3)}=-\frac16\cdot\frac32\zeta(3) \\ &=-\frac14\zeta(3)\qquad\cdots(6) \end{aligned}


    이제 (2)(2)~(6)(6)의 값을 (1)(1)에 대입하고 정리하면 최종적으로 Li3(1/2)\mathrm{Li}_3(1/2)의 값을 구할 수 있다.

    Li3 ⁣(12)=12∫01ln⁡22−2ln⁡2ln⁡(1+t)+ln⁡2t+ln⁡2(1+t)+2ln⁡2ln⁡t−2ln⁡tln⁡(1+t)1+t dt=12[0+32ζ(3)+13ln⁡32+(−π26ln⁡2)−(−14ζ(3))]=16ln⁡32−π212ln⁡2+78ζ(3)\displaystyle \begin{aligned} \mathrm{Li}_3\!\left(\dfrac12\right)&=\frac12\int_0^1\frac{{\color{blue}\ln^22-2\ln2\ln{(1+t)}}+{\color{red}\ln^2t}+{\color{limegreen}\ln^2(1+t)}+{\color{darkorchid}2\ln2\ln t}-{\color{fuchsia}2\ln t\ln{(1+t)} }}{1+t}\,\mathrm{d}t \\ &=\frac12\left[{\color{blue}0}+{\color{red}\frac32\zeta(3)}+{\color{limegreen}\frac13\ln^32}+\left({\color{darkorchid}-\frac{\pi^2}6\ln2}\right)-\left({\color{fuchsia}-\frac14\zeta(3)}\right)\right] \\ &=\frac16\ln^32-\frac{\pi^2}{12}\ln2+\frac78\zeta(3) \end{aligned}

5. 관련 문서[편집]

[1] ℜ(s)>1\Re(s) > 1인 복소수에 대해서 Lis(1)\mathrm{Li}_s (1 )은 리만 제타 함수와 동치이다.