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Question Number 135055 by mnjuly1970 last updated on 09/Mar/21
            ....dilogarithm   integral....           calculate:::               𝛗=∫_0 ^( 1) li_2 (1βˆ’x^2 )dx=?
$$\:\:\:\:\:\:\:\:\:\:\:\:….{dilogarithm}\:\:\:{integral}…. \\ $$$$\:\:\:\:\:\:\:\:\:{calculate}::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {li}_{\mathrm{2}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right){dx}=? \\ $$$$ \\ $$$$ \\ $$
Answered by Ñï= last updated on 10/Mar/21
∫_0 ^1 Li_2 (1βˆ’x^2 )dx=xLi_2 (1βˆ’x^2 )∣_0 ^1 βˆ’βˆ«_0 ^1 4x^2 βˆ™((lnx)/(1βˆ’x^2 ))dx  =4∫_0 ^1 (1βˆ’(1/(1βˆ’x^2 )))lnxdx=4∫_0 ^1 lnxdxβˆ’4∫_0 ^1 ((lnx)/(1βˆ’x^2 ))dx  =4(xlnxβˆ’x)∣_0 ^1 βˆ’4[(1/2)ln((1+x)/(1βˆ’x))lnx∣_0 ^1 βˆ’(1/2)∫_0 ^1 ((ln(1+x)βˆ’ln(1βˆ’x))/x)dx]  =βˆ’4+2∫_0 ^1 ((ln(1+x)βˆ’ln(1βˆ’x))/x)dx  =βˆ’4+2[βˆ’Li_2 (βˆ’1)+Li_2 (1)]  =βˆ’4+3Li_2 (1)  =(Ο€^2 /2)βˆ’4
$$\int_{\mathrm{0}} ^{\mathrm{1}} {Li}_{\mathrm{2}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right){dx}={xLi}_{\mathrm{2}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \centerdot\frac{{lnx}}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }\right){lnxdx}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {lnxdx}βˆ’\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\left({xlnx}βˆ’{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} βˆ’\mathrm{4}\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\frac{\mathrm{1}+{x}}{\mathrm{1}βˆ’{x}}{lnx}\mid_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)βˆ’{ln}\left(\mathrm{1}βˆ’{x}\right)}{{x}}{dx}\right] \\ $$$$=βˆ’\mathrm{4}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)βˆ’{ln}\left(\mathrm{1}βˆ’{x}\right)}{{x}}{dx} \\ $$$$=βˆ’\mathrm{4}+\mathrm{2}\left[βˆ’{Li}_{\mathrm{2}} \left(βˆ’\mathrm{1}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}\right)\right] \\ $$$$=βˆ’\mathrm{4}+\mathrm{3}{Li}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}βˆ’\mathrm{4} \\ $$
Commented by mnjuly1970 last updated on 10/Mar/21
 very nice very nice   with the thanking...
$$\:{very}\:{nice}\:{very}\:{nice} \\ $$$$\:{with}\:{the}\:{thanking}… \\ $$

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