Question Number 135055 by mnjuly1970 last updated on 09/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:....{dilogarithm}\:\:\:{integral}.... \\ $$$$\:\:\:\:\:\:\:\:\:{calculate}::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {li}_{\mathrm{2}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right){dx}=? \\ $$$$ \\ $$$$ \\ $$
Answered by ΓΓ―= last updated on 10/Mar/21
![β«_0 ^1 Li_2 (1βx^2 )dx=xLi_2 (1βx^2 )β£_0 ^1 ββ«_0 ^1 4x^2 β((lnx)/(1βx^2 ))dx =4β«_0 ^1 (1β(1/(1βx^2 )))lnxdx=4β«_0 ^1 lnxdxβ4β«_0 ^1 ((lnx)/(1βx^2 ))dx =4(xlnxβx)β£_0 ^1 β4[(1/2)ln((1+x)/(1βx))lnxβ£_0 ^1 β(1/2)β«_0 ^1 ((ln(1+x)βln(1βx))/x)dx] =β4+2β«_0 ^1 ((ln(1+x)βln(1βx))/x)dx =β4+2[βLi_2 (β1)+Li_2 (1)] =β4+3Li_2 (1) =(Ο^2 /2)β4](Q135107.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {Li}_{\mathrm{2}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right){dx}={xLi}_{\mathrm{2}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} β\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \centerdot\frac{{lnx}}{\mathrm{1}β{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}β\frac{\mathrm{1}}{\mathrm{1}β{x}^{\mathrm{2}} }\right){lnxdx}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {lnxdx}β\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}β{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\left({xlnx}β{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} β\mathrm{4}\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\frac{\mathrm{1}+{x}}{\mathrm{1}β{x}}{lnx}\mid_{\mathrm{0}} ^{\mathrm{1}} β\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)β{ln}\left(\mathrm{1}β{x}\right)}{{x}}{dx}\right] \\ $$$$=β\mathrm{4}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)β{ln}\left(\mathrm{1}β{x}\right)}{{x}}{dx} \\ $$$$=β\mathrm{4}+\mathrm{2}\left[β{Li}_{\mathrm{2}} \left(β\mathrm{1}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}\right)\right] \\ $$$$=β\mathrm{4}+\mathrm{3}{Li}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}β\mathrm{4} \\ $$
Commented by mnjuly1970 last updated on 10/Mar/21
$$\:{very}\:{nice}\:{very}\:{nice} \\ $$$$\:{with}\:{the}\:{thanking}... \\ $$