Question Number 216161 by Spillover last updated on 28/Jan/25
Answered by MrGaster last updated on 15/Feb/25
![∫_0 ^1 ∫_0 ^1 ((ln(xy))/(1+xy))xydxdy=∫_0 ^1 xdx∫_0 ^1 ((ln(xy))/(1+xy))dy ∫_0 ^1 ((ln(xy))/(1+xy))dx=(1/x)∫_0 ^1 ((ln(t))/(1+t))dt ∫_0 ^x ((ln(t))/(1+t))dt=∫_0 ^1 ((ln(t))/(1+t))+∫_1 ^x ((ln(t))/(1+t))dt ∫_1 ^x ((ln(t))/(1+t))dt=ln(t)ln(1+t)∣_1 ^x −∫_1 ^x ((ln(1+t))/t)dt =ln(x)ln(1+x)−∫_1 ^x ((ln(1+x))/t)dt ∫_0 ^1 ((ln(t))/(1+t))dt=∫_0 ^1 ln(t)Σ_(n=0) ^∞ (−t)^n dt=Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^n ln(t)dt =−Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 ))=(π^2 /(12)) ∫_1 ^x ((ln(1+t))/t)dt=−Li_2 (−x)−Li_2 (−1)=Li_2 (−x)+(π^2 /(12)) ∫_0 ^1 ∫_0 ^1 ((ln(xy))/(1+xy))xydxxy=∫_0 ^1 [((ln(x)ln(1+x))/x)+(π^2 /(12))−Li_2 (−x)−(π^2 /(12))]dx =∫_0 ^1 ((ln(x)ln(1+x))/x)dx=∫_0 ^1 Li_2 (−x)dx ∫_0 ^1 Li_2 (−x)dx=−(1/4)ζ(3) ∫_0 ^1 ∫_0 ^1 ((ln(xy))/(1+xy))xydxxy=(π^2 /6)−(5/4)ζ(3)+(1/4)ζ(3)=((3ζ(3)−4)/2)](https://www.tinkutara.com/question/Q216669.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({xy}\right)}{\mathrm{1}+{xy}}{xydxdy}=\int_{\mathrm{0}} ^{\mathrm{1}} {xdx}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({xy}\right)}{\mathrm{1}+{xy}}{dy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({xy}\right)}{\mathrm{1}+{xy}}{dx}=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({t}\right)}{\mathrm{1}+{t}}{dt} \\ $$$$\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left({t}\right)}{\mathrm{1}+{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({t}\right)}{\mathrm{1}+{t}}+\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{ln}\left({t}\right)}{\mathrm{1}+{t}}{dt} \\ $$$$\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{ln}\left({t}\right)}{\mathrm{1}+{t}}{dt}=\mathrm{ln}\left({t}\right)\mathrm{ln}\left(\mathrm{1}+{t}\right)\mid_{\mathrm{1}} ^{{x}} −\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{{t}}{dt} \\ $$$$=\mathrm{ln}\left({x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right)−\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{t}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({t}\right)}{\mathrm{1}+{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left({t}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{t}\right)^{{n}} {dt}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} \mathrm{ln}\left({t}\right){dt} \\ $$$$=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{{t}}{dt}=−\mathrm{Li}_{\mathrm{2}} \left(−{x}\right)−\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\mathrm{Li}_{\mathrm{2}} \left(−{x}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({xy}\right)}{\mathrm{1}+{xy}}{xydxxy}=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\mathrm{ln}\left({x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{Li}_{\mathrm{2}} \left(−{x}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{Li}_{\mathrm{2}} \left(−{x}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{Li}_{\mathrm{2}} \left(−{x}\right){dx}=−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({xy}\right)}{\mathrm{1}+{xy}}{xydxxy}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)=\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)−\mathrm{4}}{\mathrm{2}} \\ $$