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Calculus-Evaluate-0-1-log-1-x-1-x-3-dx-

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Question Number 142469 by mnjuly1970 last updated on 01/Jun/21
...... Calculus ..... Evaluate: ∫_0 ^( 1) (((log((1/x)))/(1−x)))^3 dx=??
$$\:\:\:\:\:\:\:\:\:\:\:\:……\:\:{Calculus}\:….. \\ $$$$\:\:\:\:{Evaluate}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{log}\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}−{x}}\right)^{\mathrm{3}} {dx}=?? \\ $$
Answered by mindispower last updated on 01/Jun/21
=−∫_0 ^1 (((ln(x))/(1−x)))^3 (1/((1−x)^3 ))=(d^2 /(2dx^2 )).(1/(1−x))=Σ_(k≥2) k(k−1)x^(k−2) =Σ_(k≥0) (k+1)(k+2)x^k =Σ_(k≥0) (k+1)(k+2)∫_0 ^1 −(ln(x))^3 x^k =Σ_(k≥0) (k+1)(k+2)∫_0 ^∞ t^3 e^(−(k+1)t) dt =(1/2)Σ_(k≥0) ((k+2)/((k+1)^3 ))∫_0 ^∞ t^3 e^(−t) dt =3Σ_(k≥0) ((1/((k+1)^2 ))+(1/((k+1)^3 )))=3(ζ(2)+ζ(3))
$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}}\right)^{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\frac{{d}^{\mathrm{2}} }{\mathrm{2}{dx}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}−{x}}=\underset{{k}\geqslant\mathrm{2}} {\sum}{k}\left({k}−\mathrm{1}\right){x}^{{k}−\mathrm{2}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right){x}^{{k}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} −\left({ln}\left({x}\right)\right)^{\mathrm{3}} {x}^{{k}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{3}} {e}^{−\left({k}+\mathrm{1}\right){t}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}+\mathrm{2}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{3}} {e}^{−{t}} {dt} \\ $$$$=\mathrm{3}\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }\right)=\mathrm{3}\left(\zeta\left(\mathrm{2}\right)+\zeta\left(\mathrm{3}\right)\right) \\ $$
Commented by mnjuly1970 last updated on 02/Jun/21
thank you so much mr power excellent as always...
$$\:{thank}\:{you}\:{so}\:{much}\:{mr}\:{power} \\ $$$$\:{excellent}\:{as}\:{always}… \\ $$
Commented by mindispower last updated on 02/Jun/21
pleasur sir thank you
$${pleasur}\:{sir}\:{thank}\:{you} \\ $$

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