## Thursday, October 9, 2014

### Logarithmic Integrals

• $\displaystyle\int_0^1\ln^a\!\left(\tfrac1x\right)dx=\Gamma(a+1)$

• $\displaystyle\int_0^1x^p\ln^a\!\left(\tfrac1x\right)dx=\frac{\Gamma(a+1)}{(p+1)^{a+1}}$

• $\displaystyle\int_0^1x^p\sqrt{\ln\!\left(\tfrac1x\right)}dx=\frac{\sqrt\pi}{2\,(p+1)^{3/2}}$

• $\displaystyle\int_0^1\frac{\ln(1+x)\,\ln^2x}{x\,(1-x)}dx=\frac72\zeta_3\ln2-\frac{\pi^4}{144}$

• $\displaystyle\int_0^1\frac{\ln(1+x)\ln^2x}{x\,(1-x)^2}dx=\frac{\pi^2}2\ln2-\frac{\pi^4}{144}-\frac94\zeta_3+\frac72\zeta_3\ln2$

• $\displaystyle\int_0^1\frac{\ln(1-x)\ln^2x}{x\,(1+x)}dx=4\operatorname{Li}_4\!\left(\tfrac12\right)+\frac{\ln^42}6-\frac{\pi^2}6\ln^22-\frac{\pi^4}{30}$

• $\displaystyle\int_0^1\frac{\ln(1-x)\ln^2x}{x\,(1+x)^2}dx=4\operatorname{Li}_4\!\left(\tfrac12\right)+\frac{\ln^42}6-\frac{\pi^2}6\ln^22-\frac{\pi^4}{30}-\frac{\pi^2}2\ln2+3\zeta_3-\frac{\pi^4}{30}$

• $\displaystyle\int_0^1\frac{\ln^2(1-x)\ln x}{x\,(1+x)}dx=6\operatorname{Li}_4\!\left(\tfrac12\right)+\frac{\ln^42}4-\frac{13\pi^4}{360}$

• $\displaystyle\int_0^1\frac{\ln^2(1-x)\ln x}{x^2\,(1+x)}dx=\frac{\pi^2}3+\frac{13\pi^4}{360}-\frac{\ln^42}4-6\operatorname{Li}_4\!\left(\tfrac12\right)-4\zeta_3$

• $\displaystyle\int_0^1\frac{\ln^2(1+x)\,\ln(1-x)}xdx=-\frac{\pi^4}{240}$

• $\displaystyle\int_0^1\frac{\ln(1+x)\,\ln^2(1-x)}xdx=2\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{\pi^4}{144}-\frac{\pi^2}{12}\ln^22+\frac{\ln^42}{12}+\frac{7\,\zeta_3}4\ln2$

• $\displaystyle\int_0^1\frac{\ln x\,\ln(1+x)\,\ln(1-x)}xdx=2\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{3\pi^4}{160}-\frac{\pi^2}{12}\ln^22+\frac{\ln^42}{12}+\frac{7\,\zeta_3}4\ln2$