Question Number 162002 by mnjuly1970 last updated on 25/Dec/21
$$ \\ $$$$\:\:\:\:\:\:\:\:{calculate} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{Li}_{\:\mathrm{2}} \:\left(\mathrm{1}\:−\:{x}^{\:\mathrm{4}} \right){dx}\:=\:? \\ $$$$\:\:\:\:−−−−− \\ $$
Answered by Lordose last updated on 27/Dec/21
$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{4}} \right)\mathrm{dx} \\ $$$$\Omega\:\overset{\mathrm{x}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} } {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$$$\Omega\:\overset{\boldsymbol{\mathrm{IBP}}} {=}\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{4x}^{\frac{\mathrm{1}}{\mathrm{4}}} \boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \:−\:\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{log}\left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{x}}\mathrm{dx}\right) \\ $$$$\Omega\:=\:−\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{x}^{\mathrm{n}+\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}} \mathrm{log}\left(\mathrm{x}\right)\:\overset{\boldsymbol{\mathrm{IBP}}×\mathrm{2}} {=}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{N}}.\boldsymbol{\mathrm{B}}\:::\:\boldsymbol{\psi}^{\left(\boldsymbol{\mathrm{m}}\right)} \left(\boldsymbol{\mathrm{z}}\right)\:=\:\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{m}}+\mathrm{1}} \boldsymbol{\mathrm{m}}!\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{z}}+\boldsymbol{\mathrm{k}}\right)^{\boldsymbol{\mathrm{m}}+\mathrm{1}} }\: \\ $$$$\boldsymbol{\Omega}\:=\:\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{5}}{\mathrm{4}}\right)\:=\:\mathrm{8}\boldsymbol{\mathrm{G}}\:+\:\boldsymbol{\pi}^{\mathrm{2}} \:−\:\mathrm{16} \\ $$$$\boldsymbol{\varnothing\mathrm{sE}} \\ $$