7ceb877eb78a60ff36e998bf06d4231d99ba0d81163e6c7b69ac7c

\displaystyle \int_{-\infty}^0 \mathrm{Ei}^3(x) \,\mathrm{d}x = -3\,\mathrm{Li}_2\biggl(\frac14\biggr) -6\ln^22 \approx -3.6856760008∫−∞0Ei3(x)dx=−3Li2(41)−6ln22≈−3.6856760008