Question Number 120774 by mnjuly1970 last updated on 02/Nov/20
$$\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right)}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}\overset{???} {=}−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\left({ln}\left(\mathrm{3}\right)+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}.\mathrm{1970}… \\ $$
Answered by mindispower last updated on 02/Nov/20
![let x^3 =t⇒dx=t^(−2(1/3)) (dt/3) ⇒Ω=(1/3)∫_0 ^1 ((ln(t^(1/3) ))/((1−t)^(1/3) )).t^(−2(1/3)) (dt/3) Ω=(1/9)∫_0 ^1 (1−t)^(−(1/3)) t^(−2(1/3)) ln(t)dt ln(t)=(∂/∂a)x^a ∣_(x=0) Ω=∂_a (1/9)∫_0 ^1 (1−t)^(1/3) t^(−(2/3)+a) dt 9Ω=(∂/∂a)β((2/3),(1/3)+a)∣_(a=0) (∂/∂y)β(x,y)=β(x,y)[Ψ(y)−Ψ(x+y)] (∂β/∂a)=β((2/3) ,(1/3)+a)[Ψ((1/3) +a)−Ψ(1+a)] 9Ω=β((1/3),(2/3))[Ψ((1/3))−Ψ(1)] Gausse theorem Ψ((p/q))=−γ−ln(2q)−(π/2)cot(((pπ)/q))+2Σ_(n=1) ^([((q−1)/2)]) cos(((2πnp)/q))ln(sin(((nπ)/q))) Ψ((1/3))=−γ−ln(6)−(π/2).(1/( (√3)))−ln(((√3)/2)) =−γ−ln(3(√3))−(π/(2(√3))) Ψ(1)=−γ β((1/3),(2/3))=((Γ((1/3))Γ((2/3)))/(Γ(2)))=.Γ((1/3))Γ((2/3))=(π/(sin((π/3)))) =((2π)/( (√3))) 9Ω=((2π)/( (√3)))(−γ−ln(3(√3))−(π/(2(√3)))+γ) =((2π)/( (√3)))(−ln(3(√3))−(π/(2(√3)))) Ω=−(π/(27))(−(√3)ln(27)−π)=−(π/(27))(π+(√3) ln(27))](https://www.tinkutara.com/question/Q120792.png)
$${let}\:{x}^{\mathrm{3}} ={t}\Rightarrow{dx}={t}^{−\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}} \frac{{dt}}{\mathrm{3}} \\ $$$$\Rightarrow\Omega=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)}{\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }.{t}^{−\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}} \frac{{dt}}{\mathrm{3}} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {t}^{−\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}} {ln}\left({t}\right){dt} \\ $$$${ln}\left({t}\right)=\frac{\partial}{\partial{a}}{x}^{{a}} \mid_{{x}=\mathrm{0}} \\ $$$$\Omega=\partial_{{a}} \frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {t}^{−\frac{\mathrm{2}}{\mathrm{3}}+{a}} {dt} \\ $$$$\mathrm{9}\Omega=\frac{\partial}{\partial{a}}\beta\left(\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}+{a}\right)\mid_{{a}=\mathrm{0}} \\ $$$$\frac{\partial}{\partial{y}}\beta\left({x},{y}\right)=\beta\left({x},{y}\right)\left[\Psi\left({y}\right)−\Psi\left({x}+{y}\right)\right] \\ $$$$\frac{\partial\beta}{\partial{a}}=\beta\left(\frac{\mathrm{2}}{\mathrm{3}}\:,\frac{\mathrm{1}}{\mathrm{3}}+{a}\right)\left[\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\:+{a}\right)−\Psi\left(\mathrm{1}+{a}\right)\right] \\ $$$$\mathrm{9}\Omega=\beta\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)\left[\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\Psi\left(\mathrm{1}\right)\right] \\ $$$${Gausse}\:{theorem} \\ $$$$\Psi\left(\frac{{p}}{{q}}\right)=−\gamma−{ln}\left(\mathrm{2}{q}\right)−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{{p}\pi}{{q}}\right)+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\left[\frac{{q}−\mathrm{1}}{\mathrm{2}}\right]} {\sum}}{cos}\left(\frac{\mathrm{2}\pi{np}}{{q}}\right){ln}\left({sin}\left(\frac{{n}\pi}{{q}}\right)\right) \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=−\gamma−{ln}\left(\mathrm{6}\right)−\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−{ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=−\gamma−{ln}\left(\mathrm{3}\sqrt{\mathrm{3}}\right)−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\Psi\left(\mathrm{1}\right)=−\gamma \\ $$$$\beta\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\mathrm{2}\right)}=.\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{9}\Omega=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\left(−\gamma−{ln}\left(\mathrm{3}\sqrt{\mathrm{3}}\right)−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}+\gamma\right) \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\left(−{ln}\left(\mathrm{3}\sqrt{\mathrm{3}}\right)−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\right) \\ $$$$\Omega=−\frac{\pi}{\mathrm{27}}\left(−\sqrt{\mathrm{3}}{ln}\left(\mathrm{27}\right)−\pi\right)=−\frac{\pi}{\mathrm{27}}\left(\pi+\sqrt{\mathrm{3}}\:{ln}\left(\mathrm{27}\right)\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/Nov/20
$${thanks}\:{alot}\:{sir}\:{mindspower} \\ $$$$\:{peace}\:\:{be}\:{upon}\:{you}… \\ $$
Commented by mindispower last updated on 03/Nov/20
$${withe}\:\:{pleasur}\:{sir}\:{have}\:{a}\:{nice}\:{Day} \\ $$