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Question Number 216161 by Spillover last updated on 28/Jan/25

Answered by MrGaster last updated on 15/Feb/25

$${\int }_0^1{\int }_0^1\frac{\ln \left(xy\right)}{1+xy}xydxdy={\int }_0^{1}xdx{\int }_0^1\frac{\ln \left(xy\right)}{1+xy}dy$$$${\int }_0^1\frac{\ln \left(xy\right)}{1+xy}dx=\frac{1}{x}{\int }_0^1\frac{\ln \left(t\right)}{1+t}dt$$$${\int }_0^x\frac{\ln \left(t\right)}{1+t}dt={\int }_0^1\frac{\ln \left(t\right)}{1+t}+{\int }_1^x\frac{\ln \left(t\right)}{1+t}dt$$$${\int }_1^x\frac{\ln \left(t\right)}{1+t}dt=\ln \left(t\right)\ln (1+t){\mid }_1^x-{\int }_1^x\frac{\ln (1+t)}{t}dt$$$$=\ln \left(x\right)\ln (1+x)-{\int }_1^x\frac{\ln (1+x)}{t}dt$$$${\int }_0^1\frac{\ln \left(t\right)}{1+t}dt={\int }_0^1\ln \left(t\right)\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-t)}^{n}dt=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^1{t}^n\ln \left(t\right)dt$$$$=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(n+1)}^2}=\frac{{\pi }^2}{12}$$$${\int }_1^x\frac{\ln (1+t)}{t}dt=-{\mathrm{Li}}_2(-x)-{\mathrm{Li}}_2(-1)={\mathrm{Li}}_2(-x)+\frac{{\pi }^2}{12}$$$${\int }_0^1{\int }_0^1\frac{\ln \left(xy\right)}{1+xy}xydxxy={\int }_0^1[\frac{\ln \left(x\right)\ln (1+x)}{x}+\frac{{\pi }^2}{12}-{\mathrm{Li}}_2(-x)-\frac{{\pi }^2}{12}]dx$$$$={\int }_0^1\frac{\ln \left(x\right)\ln (1+x)}{x}dx={\int }_0^1{\mathrm{Li}}_2(-x)dx$$$${\int }_0^1{\mathrm{Li}}_2(-x)dx=-\frac{1}{4}\zeta \left(3\right)$$$${\int }_0^1{\int }_0^1\frac{\ln \left(xy\right)}{1+xy}xydxxy=\frac{{\pi }^2}{6}-\frac{5}{4}\zeta \left(3\right)+\frac{1}{4}\zeta \left(3\right)=\frac{3\zeta \left(3\right)-4}{2}$$

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