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Tables of Generalized Logarithmic Integrals


Shobhit Site Admin
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We denote the Generalized Logarithmic Integral by

\(\displaystyle L\left[ \begin{matrix} a,b,c \\ d,e,f \end{matrix};z\right] =\int_0^z \frac{\log^a x \log^b(1-x)\log^c(1+x)}{x^d (1-x)^e (1+x)^f} dx \tag{1}\)




\((1)\) has been evaluated for certain \(a,b,c,d,e,f\) and \(z\). Here is a list of few such values:

1. \(\displaystyle L\left[ \begin{matrix} 0,n,0 \\ 1,0,0\end{matrix};z\right]=(-1)^n n! \left(\zeta(n+1)-\text{Li}_{n+1}(1-z) \right)+n! \sum_{j=1}^n \frac{(-1)^{j}\log^{n-j+1}(1-z)}{(n-j+1)!}\text{Li}_j(1-z)\)

2. \(\displaystyle L\left[ \begin{matrix} 0,0,n \\ 1,0,0\end{matrix};z\right]=\frac{\log^{n+1}(1+z)}{1+n}-n! \sum_{j=1}^n \frac{\log^{n-j+1}(1+z)}{(n-j+1)!}\text{Li}_j \left(\frac{1}{1+z} \right)+n! \zeta(n+1)-n! \text{Li}_{n+1} \left( \frac{1}{1+z}\right)\)

3. \(\displaystyle L\left[ \begin{matrix} 0,1,1 \\ 1,0,0\end{matrix};1\right]=- \frac{5\zeta(3)}{8}\)

4. \(\displaystyle L\left[ \begin{matrix} 0,1,1 \\ 2,0,0\end{matrix};1\right]=-\frac{\pi ^2}{12}-\log ^2(2)\)

5. \(\displaystyle L\left[ \begin{matrix} 0,2,1 \\ 0,0,0\end{matrix};1\right]=\frac{7 \zeta (3)}{2}-6+\frac{2}{3} (\log (2)-3) \log ^2(2) -\frac{1}{6} \pi ^2 (\log (4)-2)+\log (16)\)

6. \(\displaystyle L\left[ \begin{matrix} 1,0,1 \\ 0,1,0\end{matrix};1\right]= \zeta(3)-\frac{\pi^2}{4}\log(2)\)

7. \(\displaystyle L\left[ \begin{matrix} 1,1,1 \\ 0,0,0\end{matrix};1\right]= -6 + 4 \log 2 - \log^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \log 2 + \frac{21}{8} \zeta(3)\)

8. \(\displaystyle L\left[ \begin{matrix} 1,1,1 \\ 1,0,0\end{matrix};1\right]= -\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12}+2 \text{Li}_4 \left(\frac{1}{2} \right)\)

9. \(\displaystyle L\left[ \begin{matrix} 1,1,1 \\ 0,1,0\end{matrix};1\right]= \frac{17 \pi^4}{1440}-\frac{\pi^2}{24}\log^2(2)-\frac{\log^4(2)}{12}+\frac{7}{8}\zeta(3)\log(2)-2 \text{Li}_4 \left( \frac{1}{2}\right)\)

10. \(\displaystyle L\left[ \begin{matrix} 1,1,2 \\ 1,0,0\end{matrix};1\right]= \frac{7}{8}\zeta(2)\zeta(3) - \frac{25}{16} \zeta(5)\)

If you know any more values, please PM them to me.

Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

11. \(\displaystyle \displaystyle L\left[ \begin{matrix} 2,1,0 \\ 1,0,0\end{matrix};z\right]= 2\log( z) \text{Li}_3(z)-\log^2( z) \text{Li}_2(z)-2\text{Li}_4(z)\)

12. \(\displaystyle \displaystyle L\left[ \begin{matrix} 2,1,0 \\ 0,1,0\end{matrix};x\right]= -2\left[ \text{Li}_4 \left( \frac{-x}{1-x}\right)+\text{Li}_4(x)-\text{Li}_4(1-x)+\text{Li}_4(1)\right]+2 \left[\log(1-x)\text{Li}_3(x)-\log x \text{Li}_3(1-x) \right]\)

\(\displaystyle +2\log x \log(1-x) \text{Li}_2(1-x)-\frac{\pi^2}{6}\log^2(1-x)+\frac{1}{2}\log^2(1-x) \log^2 (x)+\frac{1}{3}\log(x)\log^3(1-x)-\frac{1}{12}\log^4(1-x)\)

\(\displaystyle +2\zeta(3) \log \left(\frac{x}{1-x} \right)\)

13. \(\displaystyle \displaystyle L\left[ \begin{matrix} 1,0,2 \\ 0,1,0\end{matrix};1\right]=-\frac{7\pi^4}{144}-\frac{5}{12}\pi^2 \log^2(2)+\frac{\log^4(2)}{2}+\frac{21}{4}\zeta(3) \log(2) +4 \text{Li}_4 \left(\frac{1}{2} \right)\)

Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

The following were proposed by DreamWeaver:

14. \(\displaystyle L\left[ \begin{matrix} m,0,0 \\ 0,1,0 \end{matrix};z\right]=m! \, \sum_{j=0}^m(-1)^j\frac{(\log z)^{m-j}}{(m-j)!}\text{Li}_{j+1}(z)\)

15. \(\displaystyle L\left[ \begin{matrix} m,0,0 \\ 0,0,1 \end{matrix};z\right]=m! \, \sum_{j=0}^m(-1)^{j+1}\frac{(\log z)^{m-j}}{(m-j)!}\text{Li}_{j+1}(-z)\)

16. \(\displaystyle L\left[ \begin{matrix} m,0,0 \\ 1,1,0 \end{matrix};z\right]= m! \, \sum_{j=0}^m(-1)^j\frac{(\log z)^{m-j}}{(m-j)!}\chi_{j+1}(z)\) where \(\chi_\nu(z)\) is the Legendre's Chi Function.

17. \(\displaystyle L\left[ \begin{matrix} m,0,0 \\ \tfrac{1}{2},0,1 \end{matrix};z^2\right]=2^{m+1}m! \, \sum_{j=0}^m(-1)^j\frac{(\log z)^{m-j}}{(m-j)!}\text{Ti}_{j+1}(z)\) where \(\text{Ti}_\nu(z)\) is the Inverse Tangent Integral.

18. \(\displaystyle L\left[ \begin{matrix} 0,m,0 \\ 0,0,1 \end{matrix};z\right]=\log^m(1-z)\log\left( \frac{1+z}{2} \right)+(-1)^{m+1}m! \, \text{Li}_{m+1}\left(\frac{1}{2}\right)+\)

\(\displaystyle m! \, \sum_{j=1}^m(-1)^{j+1} \frac{\log^{m-j}(1-z)}{(m-j)!}\text{Li}_{j+1}\left(\frac{1-z}{2}\right)\)

19. \(\displaystyle L\left[ \begin{matrix} 0,0,m \\ 0,1,0 \end{matrix};z\right]= -\log^m(1+z)\log\left( \frac{1-z}{2} \right)+(-1)^{m+1}m! \, \text{Li}_{m+1}\left(\frac{1}{2}\right)+\)
\(\displaystyle m! \, \sum_{j=1}^m(-1)^{j} \frac{\log^{m-j}(1+z)}{(m-j)!}\text{Li}_{j+1}\left(\frac{1+z}{2}\right)\)

Shobhit Site Admin
Site Admin

Posts: 852
Location: Jaipur, India

Some High Order Logarithmic Integrals:

20. \(\displaystyle L\left[ \begin{matrix} 2,3,0 \\ 1,0,0 \end{matrix};1\right]=-\frac{23}{1260}\pi^6+12\zeta^2(3)\)

21. \(\displaystyle L\left[ \begin{matrix} 4,3,0 \\ 1,0,0 \end{matrix};1\right]=-\frac{61 \pi^8}{1575}-12\pi^2 \zeta^2(3)+432\zeta(3)\zeta(5)\)

22. \(\displaystyle L\left[ \begin{matrix} 0,2,1 \\ 0,0,1 \end{matrix};1\right]=\displaystyle 2\zeta (3) \log 2-\frac{\pi ^4}{360}+\frac{ \log ^4 2}{4}-\frac{1}{6} \pi ^2 \log ^2 2\)


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