Menu Close

dilogarithm-integral-calculate-0-1-li-2-1-x-2-dx-




Question Number 135055 by mnjuly1970 last updated on 09/Mar/21
            ....dilogarithm   integral....           calculate:::               𝛗=∫_0 ^( 1) li_2 (1−x^2 )dx=?
.dilogarithmintegral.calculate:::\boldsymbolϕ=01li2(1x2)dx=?
Answered by Ñï= last updated on 10/Mar/21
∫_0 ^1 Li_2 (1−x^2 )dx=xLi_2 (1−x^2 )∣_0 ^1 −∫_0 ^1 4x^2 ∙((lnx)/(1−x^2 ))dx  =4∫_0 ^1 (1−(1/(1−x^2 )))lnxdx=4∫_0 ^1 lnxdx−4∫_0 ^1 ((lnx)/(1−x^2 ))dx  =4(xlnx−x)∣_0 ^1 −4[(1/2)ln((1+x)/(1−x))lnx∣_0 ^1 −(1/2)∫_0 ^1 ((ln(1+x)−ln(1−x))/x)dx]  =−4+2∫_0 ^1 ((ln(1+x)−ln(1−x))/x)dx  =−4+2[−Li_2 (−1)+Li_2 (1)]  =−4+3Li_2 (1)  =(π^2 /2)−4
01Li2(1x2)dx=xLi2(1x2)01014x2lnx1x2dx=401(111x2)lnxdx=401lnxdx401lnx1x2dx=4(xlnxx)014[12ln1+x1xlnx011201ln(1+x)ln(1x)xdx]=4+201ln(1+x)ln(1x)xdx=4+2[Li2(1)+Li2(1)]=4+3Li2(1)=π224
Commented by mnjuly1970 last updated on 10/Mar/21
 very nice very nice   with the thanking...
veryniceverynicewiththethanking

Leave a Reply

Your email address will not be published. Required fields are marked *