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Question Number 172953 by Mathspace last updated on 03/Jul/22
find ∫_0 ^1 (√(1−x^4 ))lnx dx
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{lnx}\:{dx} \\ $$
Answered by Ar Brandon last updated on 03/Jul/22
Ω=∫_0 ^1 (√(1−x^4 ))lnxdx , x=t^(1/4) =(1/(16))∫_0 ^1 t^(−(3/4)) (√(1−t))lntdt=(1/(16))∙(∂/∂α)∣_(α=(1/4)) ∫_0 ^1 t^(α−1) (√(1−t))dt =(1/(16))∙(∂/∂α)∣_(α=(1/4)) β(α, (3/2))=(1/(16))∙(∂/∂α) ((Γ(α)Γ((1/2))(1/2))/(Γ(α+(3/2)))) _(α=(1/4)) =((√π)/(32))∙(((Γ′(α))/(Γ(α+(3/2))))−((Γ(α)ψ(α+(3/2)))/(Γ(α+(3/2)))))_(α=(1/4)) =((√π)/(32))(((Γ((1/4))ψ((1/4))−Γ((1/4))ψ((7/4)))/(Γ((7/4))))) =((√π)/(32))∙((Γ((1/4)))/((3/4)Γ((3/4))))(ψ((1/4))−ψ((3/4))−(4/3)) =((√π)/(24))∙((Γ^2 ((1/4)))/(π(√2)))(−πcot((π/4))−(4/3))=−((√(2π))/(48π))(((3π+4)/3))Γ^2 ((1/4))
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }\mathrm{ln}{xdx}\:,\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{4}}} \: \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{−\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{1}−{t}}\mathrm{ln}{tdt}=\frac{\mathrm{1}}{\mathrm{16}}\centerdot\frac{\partial}{\partial\alpha}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\alpha−\mathrm{1}} \sqrt{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\centerdot\frac{\partial}{\partial\alpha}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{4}}} \beta\left(\alpha,\:\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{16}}\centerdot\frac{\partial}{\partial\alpha}\:\frac{\Gamma\left(\alpha\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{\mathrm{1}}{\mathrm{2}}}{\Gamma\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:_{\alpha=\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{32}}\centerdot\left(\frac{\Gamma'\left(\alpha\right)}{\Gamma\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right)}−\frac{\Gamma\left(\alpha\right)\psi\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right)}\right)_{\alpha=\frac{\mathrm{1}}{\mathrm{4}}} =\frac{\sqrt{\pi}}{\mathrm{32}}\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\psi\left(\frac{\mathrm{7}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{7}}{\mathrm{4}}\right)}\right) \\ $$$$\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{32}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\frac{\mathrm{3}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{24}}\centerdot\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\pi\sqrt{\mathrm{2}}}\left(−\pi\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{4}}{\mathrm{3}}\right)=−\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{48}\pi}\left(\frac{\mathrm{3}\pi+\mathrm{4}}{\mathrm{3}}\right)\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$
Commented by Ar Brandon last updated on 04/Jul/22
Commented by Ar Brandon last updated on 04/Jul/22
Commented by Tawa11 last updated on 06/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by Mathspace last updated on 04/Jul/22
Υ=∫_0 ^1 (√(1−x^4 ))lnx dx changement x^4 =t give Υ=∫_0 ^1 (√(1−t))ln(t^(1/4) ).(1/4)t^((1/4)−1) dt =(1/(16))∫_0 ^1 t^((1/4)−1) (1−t)^(1/2) lnt dt let f(a)=∫_0 ^1 t^(a−1) (1−t)^(1/2) dt ⇒ f^′ (a)=∫_0 ^1 t^(a−1) (1−t)^(1/2) lnt ⇒ ∫_0 ^1 t^((1/4)−1) (1−t)^(1/2) lnt dt=f^′ ((1/4)) we have f(a)=B(a,(3/2)) =((Γ(a).Γ((3/2)))/(Γ(a+(3/2)))) Γ((3/2))=Γ((1/2)+1)=(1/2)Γ((1/2))=((√π)/2) Γ(a+(3/2))=Γ(a+(1/2)+1) =(a+(1/2))Γ(a+(1/2)) f(a)=((√π)/2)×((Γ(a))/(Γ(a+(3/2)))) ⇒ f^′ (a)=((√π)/2).((Γ^′ (a)Γ(a+(3/2))−Γ(a)Γ^′ (a+(3/2)))/(Γ^2 (a+(3/2)))) ⇒f^′ ((1/4))=((√π)/2).((Γ^′ ((1/4))Γ((1/4)+(3/2))−Γ((1/4))Γ^′ ((1/4)+(3/2)))/(Γ^2 ((1/4)+(3/2)))) =((√π)/2).((Γ^′ ((1/4))Γ((7/4))−Γ((1/4))Γ^′ ((7/4)))/(Γ^2 ((7/4)))) and Υ=(1/(16))f^′ ((1/4))
$$\Upsilon=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{lnx}\:{dx}\:{changement} \\ $$$${x}^{\mathrm{4}} ={t}\:{give} \\ $$$$\Upsilon=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}}{ln}\left({t}^{\frac{\mathrm{1}}{\mathrm{4}}} \right).\frac{\mathrm{1}}{\mathrm{4}}{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {lnt}\:{dt} \\ $$$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dt}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {lnt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {lnt}\:{dt}={f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${we}\:{have}\:{f}\left({a}\right)={B}\left({a},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\Gamma\left({a}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\Gamma\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${f}\left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\Gamma\left({a}\right)}{\Gamma\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma^{'} \left({a}\right)\Gamma\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)−\Gamma\left({a}\right)\Gamma^{'} \left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{4}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma^{'} \left(\frac{\mathrm{7}}{\mathrm{4}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{7}}{\mathrm{4}}\right)} \\ $$$${and}\:\Upsilon=\frac{\mathrm{1}}{\mathrm{16}}{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$
Commented by Tawa11 last updated on 06/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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