一些推广的对数积分的特殊值
首先我们定义
(1)中 \(a,b,c,d,e,f\) 和 \(z\) 为整数.
下面是一些特殊值
\(\displaystyle L\left[ \begin{matrix} 0,n,0 \\ 1,0,0\end{matrix};z\right]=(-1)^n n! \left(\zeta(n+1)-\text{Li}_{n+1}(1-z) \right)+n! \sum_{j=1}^n \frac{(-1)^{j}\ln^{n-j+1}(1-z)}{(n-j+1)!}\text{Li}_j(1-z)\)
\(\displaystyle L\left[ \begin{matrix} 0,0,n \\ 1,0,0\end{matrix};z\right]\)
\(\displaystyle =\frac{\ln^{n+1}(1+z)}{1+n}-n! \sum_{j=1}^n \frac{\ln^{n-j+1}(1+z)}{(n-j+1)!}\text{Li}_j \left(\frac{1}{1+z} \right)+n! \zeta(n+1)-n! \text{Li}_{n+1} \left( \frac{1}{1+z}\right)\)
\(\displaystyle L\left[ \begin{matrix} 0,1,1 \\ 1,0,0\end{matrix};1\right]=- \frac{5\zeta(3)}{8}\)
\(\displaystyle L\left[ \begin{matrix} 0,1,1 \\ 2,0,0\end{matrix};1\right]=-\frac{\pi ^2}{12}-\ln ^2(2)\)
\(\displaystyle L\left[ \begin{matrix} 0,2,1 \\ 0,0,0\end{matrix};1\right]=\frac{7 \zeta (3)}{2}-6+\frac{2}{3} (\ln (2)-3) \ln ^2(2) -\frac{1}{6} \pi ^2 (\ln (4)-2)+\ln (16)\)
\(\displaystyle L\left[ \begin{matrix} 1,0,1 \\ 0,1,0\end{matrix};1\right]= \zeta(3)-\frac{\pi^2}{4}\ln(2)\)
\(\displaystyle L\left[ \begin{matrix} 1,1,1 \\ 0,0,0\end{matrix};1\right]= -6 + 4 \ln 2 - \ln^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \ln 2 + \frac{21}{8} \zeta(3)\)
\(\displaystyle L\left[ \begin{matrix} 1,1,1 \\ 1,0,0\end{matrix};1\right]= -\frac{3 \pi^4}{160}+\frac{7\ln(2)}{4}\zeta(3)-\frac{\pi^2 \ln^2(2)}{12} +\frac{\ln^4(2)}{12}+2 \text{Li}_4 \left(\frac{1}{2} \right)\)
\(\displaystyle L\left[ \begin{matrix} 1,1,1 \\ 0,1,0\end{matrix};1\right]= \frac{17 \pi^4}{1440}-\frac{\pi^2}{24}\ln^2(2)-\frac{\ln^4(2)}{12}+\frac{7}{8}\zeta(3)\ln(2)-2 \text{Li}_4 \left( \frac{1}{2}\right)\)
\(\displaystyle L\left[ \begin{matrix} 1,1,2 \\ 1,0,0\end{matrix};1\right]= \frac{7}{8}\zeta(2)\zeta(3) - \frac{25}{16} \zeta(5)\)
\(\displaystyle L\left[ \begin{matrix} 2,1,0 \\ 1,0,0\end{matrix};z\right]= 2\ln( z) \text{Li}_3(z)-\ln^2( z) \text{Li}_2(z)-2\text{Li}_4(z)\)
\(\displaystyle L\left[ \begin{matrix} 2,1,0 \\ 0,1,0\end{matrix};x\right]= -2\left[ \text{Li}_4 \left( \frac{-x}{1-x}\right)+\text{Li}_4(x)-\text{Li}_4(1-x)+\text{Li}_4(1)\right]+2 \Big[\ln(1-x)\text{Li}_3(x)\)
\(\displaystyle -\ln x \text{Li}_3(1-x) \Big]+2\ln x \ln(1-x) \text{Li}_2(1-x)-\frac{\pi^2}{6}\ln^2(1-x)+\frac{1}{2}\ln^2(1-x) \ln^2 (x)\)
\(\displaystyle +\frac{1}{3}\ln(x)\ln^3(1-x)-\frac{1}{12}\ln^4(1-x)+2\zeta(3) \ln \left(\frac{x}{1-x} \right)\)
\(\displaystyle L\left[ \begin{matrix} 1,0,2 \\ 0,1,0\end{matrix};1\right]=-\frac{7\pi^4}{144}-\frac{5}{12}\pi^2 \ln^2(2)+\frac{\ln^4(2)}{2}+\frac{21}{4}\zeta(3) \ln(2) +4 \text{Li}_4 \left(\frac{1}{2} \right)\)
\(\displaystyle L\left[ \begin{matrix} m,0,0 \\ 0,1,0 \end{matrix};z\right]=m! \, \sum_{j=0}^m(-1)^j\frac{(\ln z)^{m-j}}{(m-j)!}\text{Li}_{j+1}(z)\)
\(\displaystyle L\left[ \begin{matrix} m,0,0 \\ 0,0,1 \end{matrix};z\right]=m! \, \sum_{j=0}^m(-1)^{j+1}\frac{(\ln z)^{m-j}}{(m-j)!}\text{Li}_{j+1}(-z)\)
\(\displaystyle L\left[ \begin{matrix} m,0,0 \\ 1,1,0 \end{matrix};z\right]= m! \, \sum_{j=0}^m(-1)^j\frac{(\ln z)^{m-j}}{(m-j)!}\chi_{j+1}(z)\)
其中 \(\chi_\nu(z)\) 为 Legendre's Chi Function.
\(\displaystyle L\left[ \begin{matrix} m,0,0 \\ \dfrac{1}{2},0,1 \end{matrix};z^2\right]=2^{m+1}m! \, \sum_{j=0}^m(-1)^j\frac{(\ln z)^{m-j}}{(m-j)!}\text{Ti}_{j+1}(z)\)
其中 \(\text{Ti}_\nu(z)\) 为 Inverse Tangent Integral.
\(\displaystyle L\left[ \begin{matrix} 0,m,0 \\ 0,0,1 \end{matrix};z\right]=\ln^m(1-z)\ln\left( \frac{1+z}{2} \right)+(-1)^{m+1}m! \, \text{Li}_{m+1}\left(\frac{1}{2}\right)\)
\(\displaystyle + m! \, \sum_{j=1}^m(-1)^{j+1} \frac{\ln^{m-j}(1-z)}{(m-j)!}\text{Li}_{j+1}\left(\frac{1-z}{2}\right)\)
\(\displaystyle L\left[ \begin{matrix} 0,0,m \\ 0,1,0 \end{matrix};z\right]= -\ln^m(1+z)\ln\left( \frac{1-z}{2} \right)+(-1)^{m+1}m! \, \text{Li}_{m+1}\left(\frac{1}{2}\right)\)
\(\displaystyle +m! \, \sum_{j=1}^m(-1)^{j} \frac{\ln^{m-j}(1+z)}{(m-j)!}\text{Li}_{j+1}\left(\frac{1+z}{2}\right)\)
\(\displaystyle L\left[ \begin{matrix} 2,3,0 \\ 1,0,0 \end{matrix};1\right]=-\frac{23}{1260}\pi^6+12\zeta^2(3)\)
\(\displaystyle L\left[ \begin{matrix} 4,3,0 \\ 1,0,0 \end{matrix};1\right]=-\frac{61 \pi^8}{1575}-12\pi^2 \zeta^2(3)+432\zeta(3)\zeta(5)\)
\(\displaystyle L\left[ \begin{matrix} 0,2,1 \\ 0,0,1 \end{matrix};1\right]=\displaystyle 2\zeta (3) \ln 2-\frac{\pi ^4}{360}+\frac{ \ln ^4 2}{4}-\frac{1}{6} \pi ^2 \ln ^2 2\)
其中的一些证明可以看这里
