Monday, October 13, 2014

Table of Contents

Notations


  • $\ln x$ — the natural logarithm, also denoted in many sources as $\log x$.
  • Direct trigonometric functions are denoted as $\sin x,\,\cos x,\,\tan x,\,\cot x$.
  • Inverse trigonometric functions are denoted as $\arcsin x,\,\arccos x,\,\arctan x,\,\operatorname{arccot}x$ (and never as $\tan^{-1}x$, to avoid any confusion with powers).
  • Parentheses around the argument is usually omitted if the next term is an integer, named constant, variable, radical, or function name, e.g: $\ln2,\,\ln\pi,\,\ln\sqrt{2\pi},\,\ln\sin x,\,\ln\ln x$.
  • Expressions like $\ln2\ln3$ should be understood as $(\ln(2))\cdot(\ln(3))$.
  • Expressions like $\ln^53,\,\sin^2x$ should be understood as $(\ln(2))^5,\,(\sin(x))^2$.
  • $\zeta(x)$ — the Riemann zeta function.
  • $\zeta_3,\,\zeta_5,\,$ etc. — values of $\zeta(x)$ for odd positive integer arguments. Values at even arguments are always expanded in terms of $\pi$.

Thursday, October 9, 2014

Trilogarithm Identities

  • $\displaystyle\operatorname{Li}_3\!\left(\tfrac12\right)=\frac{\ln^32}6-\frac{\pi^2}{12}\ln2+\frac78\zeta_3$
  • $\displaystyle\operatorname{Li}_3\!\left(\tfrac14\right)=\frac{\pi^2}3\ln\left(\frac29\right)+\frac43\ln^32-2\ln2\cdot\ln^23+\frac43\ln^33+\frac{15}2\zeta_3-4\operatorname{Li}_3\!\left(\tfrac13\right)-4\operatorname{Li}_3\!\left(\tfrac23\right)$
  • $\displaystyle\operatorname{Li}_3\!\left(\tfrac34\right)=\frac43\ln^32+2\ln\left(\tfrac32\right)\left(\frac{\pi^2}3+\ln2\cdot\ln3\right)-\ln^33-\frac{13}3\zeta_3+2\operatorname{Li}_3\!\left(\tfrac13\right)+4\operatorname{Li}_3\!\left(\tfrac23\right)$
  • $\displaystyle\begin{align*}\operatorname{Li}_3\!\left(\tfrac38\right)&=\frac{14}3\ln^32-\frac{\pi^2}3\ln2+\frac{4\pi^2}3\ln3-\frac{2\pi^2}3\ln5\\&-4\ln^22\cdot\ln3+4\ln2\cdot\ln^23-2\ln^33+2\ln2\cdot\ln3\cdot\ln5-3\ln2\cdot\ln^25-\ln3\cdot\ln^25\\&+\frac43\ln^35-\frac{19}6\zeta_3+6\operatorname{Li}_3\!\left(\tfrac13\right)+7\operatorname{Li}_3\!\left(\tfrac23\right)-2\operatorname{Li}_3\!\left(\tfrac15\right)-2\operatorname{Li}_3\!\left(\tfrac25\right)-2\operatorname{Li}_3\!\left(\tfrac35\right)-2\operatorname{Li}_3\!\left(\tfrac45\right)\end{align*}$
  • $\displaystyle\begin{align*}\operatorname{Li}_3\!\left(\tfrac58\right)&=\frac{\pi^2}{12}\ln2-\frac{2\pi^2}3\ln3+\frac{\pi^2}3\ln5+\frac{23}6\ln^32-2\ln^22\cdot\ln3-5\ln2\cdot\ln^23+\frac43\ln^33\\&-\frac72\ln^22\cdot\ln5+2\ln2\cdot\ln3\cdot\ln5+\ln^23\cdot\ln5+3\ln2\cdot\ln^25-\frac76\ln^35+\frac{21}8\zeta_3\\&-6\operatorname{Li}_3\!\left(\tfrac13\right)-6\operatorname{Li}_3\!\left(\tfrac23\right)+\frac52\operatorname{Li}_3\!\left(\tfrac15\right)+3\operatorname{Li}_3\!\left(\tfrac25\right)+\frac32\operatorname{Li}_3\!\left(\tfrac45\right)+2\operatorname{Li}_3\!\left(\tfrac16\right)+2\operatorname{Li}_3\!\left(\tfrac56\right)\end{align*}$
  • More to follow...

Dilogarithm Identities


  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac12\right)=\frac{\pi^2}{12}-\frac{\ln^22}2$
  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac23\right)=\frac{\pi^2}6+\ln2\cdot\ln3-\ln^23-\operatorname{Li}_2\!\left(\tfrac13\right)$
  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac14\right)=\frac{\pi^2}6-2\ln^22+2\ln2\cdot\ln3-\ln^23-2\operatorname{Li}_2\!\left(\tfrac13\right)$
  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac34\right)=\ln^22-2\ln^22+2\operatorname{Li}_2\!\left(\tfrac13\right)$
  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac35\right)=\frac{\pi^2}6-\ln2\cdot\ln3+\ln2\cdot\ln5+\ln3\cdot\ln5-\ln^25-2\operatorname{Li}_2\!\left(\tfrac25\right)$
  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac45\right)=\frac{\pi^2}6+2\ln2\cdot\ln5-\ln^25-\operatorname{Li}_2\!\left(\tfrac15\right)$
  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac16\right)=\frac{\pi^2}{12}-\frac{\ln^22}2-2\ln2\cdot\ln3+2\ln2\cdot\ln5+\ln3\cdot\ln5-\ln^25+\operatorname{Li}_2\!\left(\tfrac13\right)-\operatorname{Li}_2\!\left(\tfrac15\right)-\operatorname{Li}_2\!\left(\tfrac25\right)$
  • $\displaystyle\operatorname{Li}_2\!\left(\tfrac56\right)=\frac{\pi^2}{12}-\frac{\ln^22}2-\ln^23-\ln2\cdot\ln5+\ln^25-\operatorname{Li}_2\!\left(\tfrac13\right)+\operatorname{Li}_2\!\left(\tfrac15\right)+\operatorname{Li}_2\!\left(\tfrac25\right)$
  • More to follow...

Logarithmic Integrals


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

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

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

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

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

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

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

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

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

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

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

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