# Question and Answers Forum

Integration Questions

Question Number 162177 by mnjuly1970 last updated on 27/Dec/21

$$\\$$$$\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}^{\:\mathrm{2}} \left(\:{x}\:\right).\:\mathrm{Li}_{\:\mathrm{2}} \:\left({x}\:\right)}{{x}^{\:} }\:{dx}\:=? \\$$

Answered by Ar Brandon last updated on 27/Dec/21

$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} {xLi}_{\mathrm{2}} \left({x}\right)}{{x}}{dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} {log}^{\mathrm{2}} {x}}{{x}}{dx} \\$$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\left[\frac{\mathrm{1}}{\mathrm{3}}{x}^{{n}} {log}^{\mathrm{3}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} β\frac{{n}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}β\mathrm{1}} {log}^{\mathrm{3}} {xdx}\right) \\$$$$\:\:\:=β\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\centerdot\frac{\partial^{\mathrm{3}} }{\partial\alpha^{\mathrm{3}} }\mid_{\alpha={n}β\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\alpha} {dx} \\$$$$\:\:\:=β\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\centerdot\frac{\partial^{\mathrm{3}} }{\partial\alpha^{\mathrm{3}} }\mid_{\alpha={n}β\mathrm{1}} \frac{\mathrm{1}}{\alpha+\mathrm{1}} \\$$$$\:\:\:=β\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left(\frac{β\mathrm{6}}{{n}^{\mathrm{4}} }\right)=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{5}} }=\mathrm{2}\zeta\left(\mathrm{5}\right) \\$$

Commented by mnjuly1970 last updated on 27/Dec/21

$$\:{grateful}\:{sir}\:{brandon} \\$$

Commented by Ar Brandon last updated on 27/Dec/21

$$\mathrm{My}\:\mathrm{pleasure},\:\mathrm{Sir}. \\$$