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먼저 다음 4가지의 참고식들을 증명한 다음, 이 참고식들을 사용하여 $$\mathrm{Li}_3\!\left(\dfrac12\right)$$의 값을 계산할 것이다.
$$n\in\mathbb{Z}^+$$, $$k\in\mathbb{Z}^+$$에 대하여 다음 두 식을 얻을 수 있다. 여기서 $$\mathbb{Z}^+$$는 양의 정수의 집합을 의미한다.
$$s\in\mathbb{Z}^+$$에 대하여 다음 식을 얻을 수 있다.
먼저 $$x=t^2$$으로 치환함으로써 시작한다.
이제 위의 참고식들을 사용하여 $$\mathrm{Li}_3\!\left(\dfrac12\right)$$의 값을 계산해보자. 계산 과정 중, 위의 참고식이나 아래의 과정에서 나타나는 관계식을 사용한 경우, 등호 위에다가 해당 참고식·관계식의 알파벳·번호를 적을 것이다. (예: $$\overset{A}=$$, $$\overset{(1)}=$$)
[math(\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.
[math(\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} )]
[math(\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} )]
[math(\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} )]
[math(\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} )]
[math(\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 $$(2)$$~$$(6)$$의 값을 $$(1)$$에 대입하고 정리하면 최종적으로 $$\mathrm{Li}_3(1/2)$$의 값을 구할 수 있다.
[math(\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} )]