Question Number 208423 by lepuissantcedricjunior last updated on 15/Jun/24
![calculons i=∫∫∫_([0;1]) ((dxdydz)/(1−xyz))](https://www.tinkutara.com/question/Q208423.png)
$$\:\:\:\boldsymbol{{calculons}}\: \\ $$$$\boldsymbol{{i}}=\int\int\int_{\left[\mathrm{0};\mathrm{1}\right]} \frac{\boldsymbol{{dxdydz}}}{\mathrm{1}−\boldsymbol{{xyz}}} \\ $$
Answered by Berbere last updated on 15/Jun/24
![=∫∫[−(1/(xy))ln(1−xy)]dydx xy=u⇒dy=(du/x) =∫_0 ^1 (1/x)∫_0 ^x ((ln(1−u))/(−u))dudx =∫_0 ^1 ((Li_2 (x))/x)dx=Li_3 (1)=ζ(3) Li_(n+1) (x)=∫_0 ^x ((Li_n (z))/z)dz Li_2 (x)=−∫_0 ^x ((ln(1−t))/t)dt Li_3 (z)=Σ_(n≥1) (z^n /n^3 );ζ(3)=Σ_(n≥1) (1/n^3 )](https://www.tinkutara.com/question/Q208435.png)
$$=\int\int\left[−\frac{\mathrm{1}}{{xy}}{ln}\left(\mathrm{1}−{xy}\right)\right]{dydx} \\ $$$${xy}={u}\Rightarrow{dy}=\frac{{du}}{{x}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} \frac{{ln}\left(\mathrm{1}−{u}\right)}{−{u}}{dudx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{{x}}{dx}={Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\zeta\left(\mathrm{3}\right) \\ $$$${Li}_{{n}+\mathrm{1}} \left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{Li}_{{n}} \left({z}\right)}{{z}}{dz} \\ $$$${Li}_{\mathrm{2}} \left({x}\right)=−\int_{\mathrm{0}} ^{{x}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$${Li}_{\mathrm{3}} \left({z}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{z}^{{n}} }{{n}^{\mathrm{3}} };\zeta\left(\mathrm{3}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$