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Question Number 139478 by mnjuly1970 last updated on 27/Apr/21
$......... nice ... ... ... calculus........ Φ:=∫_0 ^( 1) ((ln(((1+x)/2)))/(x^2 −1))dx=(π^2 /(24)) NOTE :: li_2 (z)+li_2 (1−z)=(π^2 /6)−ln(z)ln(1−z) Hence :: li_2 ((1/2))=(π^2 /(12))−(1/2)ln^2 (2) Φ:=^(⟨ ((1+x)/2)=y ⟩) 2∫_(1/2) ^( 1) ((ln(y))/(4y^2 −4y))dy :=(1/2)∫_(1/2) ^( 1) ((ln(y))/(y(y−1)))dy=(1/2) ∫_(1/2) ^( 1) {((ln(y))/(y−1))−((ln(y))/y)}dy :=−(1/2) [(1/2) ln^2 (y)]_((1 )/2) ^1 +(1/2){li_2 (1)−li_2 ((1/2))} :=(1/4)ln^2 (2)+(π^2 /(12))+(1/2)(−(π^2 /(12))−(1/2)ln^2 (2)) Φ:=(π^2 /(24))$
$. . . . . . . . . n i c e . . . . . . . . . c a l c u l u s . . . . . . . .$ $Φ := \int_{0}^{1} \frac{l n (\frac{1 + x}{2})}{x^{2} - 1} d x = \frac{π^{2}}{24}$ $N O T E :: {l i}_{2} (z) + {l i}_{2} (1 - z) = \frac{π^{2}}{6} - l n (z) l n (1 - z)$ $H e n c e :: {l i}_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{1}{2} {l n}^{2} (2)$ $Φ : \overset{⟨ \frac{1 + x}{2} = y ⟩}{=} 2 \int_{\frac{1}{2}}^{1} \frac{l n (y)}{4 y^{2} - 4 y} d y$ $:= \frac{1}{2} \int_{\frac{1}{2}}^{1} \frac{l n (y)}{y (y - 1)} d y = \frac{1}{2} \int_{\frac{1}{2}}^{1} {\frac{l n (y)}{y - 1} - \frac{l n (y)}{y}} d y$ $:= - \frac{1}{2} {[\frac{1}{2} {l n}^{2} (y)]}_{\frac{1}{2}}^{1} + \frac{1}{2} {{l i}_{2} (1) - {l i}_{2} (\frac{1}{2})}$ $:= \frac{1}{4} {l n}^{2} (2) + \frac{π^{2}}{12} + \frac{1}{2} (- \frac{π^{2}}{12} - \frac{1}{2} {l n}^{2} (2))$ $Φ := \frac{π^{2}}{24}$
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