"다이로그 함수(dilogarithm)"의 두 판 사이의 차이

2009년 7월 6일 (월) 11:45 판

\(\operatorname{Li}_2(z) = -\int_0^z{{\ln (1-t)}\over t} dt = \sum_{n=1}^\infty {z^n \over n^2}\)

\(\mbox{Li}_2(x)+\mbox{Li}_2 \left( \frac{1}{x} \right) = -\frac{\pi^2}{6}-\frac{1}{2}(\ln(-x))^2 \qquad\)

\(\mbox{Li}_2 \left(x \right)+\mbox{Li}_2 \left(1-x \right)= \frac{\pi^2}{6}-\ln(x)\ln(1-x)\)

\(\mbox{Li}_2(-x)=-\mbox{Li}_2 \left( \frac{x}{1+x} \right)-\frac{1}{2}(\ln(x+1))^2 \)

다음 값들은 알려져 있음.

\(\mbox{Li}_{2}(0)=0\)

\(\mbox{Li}_{2}(1)=\frac{\pi^2}{6}\)

\(\mbox{Li}_{2}(-1)=-\frac{\pi^2}{12}\)

\(\mbox{Li}_{2}(\frac{1}{2})=\frac{\pi^2}{12}-\frac{1}{2}\log^2(2)\)

\(\mbox{Li}_{2}(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-\log^2(\frac{1+\sqrt{5}}{2})\)

\(\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})\)

\(\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\)

\(\mbox{Li}_{2}(\frac{-1-\sqrt{5}}{2})=-\frac{\pi^2}{10}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\)

The dilogarithm is the only mathematical function with a sense of humor.

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