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Question Number 206754 by mathzup last updated on 23/Apr/24
find ∫_0 ^1 (√(1−(√x)))ln^2 (x)dx
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\sqrt{{x}}}{ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$
Answered by Berbere last updated on 24/Apr/24
x=u^2 ⇒dx=2udu ∫_0 ^1 (√(1−(√x)))ln^2 (x)dx.=A=∫_0 ^1 (1−u)^(1/2) (2ln(u))^2 .2udu =8∫_0 ^1 (1−u)^(1/2) ln^2 (u).udu f(a)=∫_0 ^1 (1−u)^((3/2)−1) u^(a−1) =β(a,(3/2))du;f′′(a)=∫_0 ^1 (1−u)^(1/2) ln^2 (u).u^(a−1) du f′(a)=β(a,(3/2))(Ψ(a)−Ψ(a+(3/2))) f′′(a)=β(a,(3/2)){(Ψ(a)−Ψ(a+(3/2)))+(Ψ′(a)−Ψ′(a+(3/2))} A=8f′′(2) =8(β(2,(3/2))){(Ψ(2)−Ψ((3/2)+2))^2 +(Ψ′(2)−Ψ′(2+(3/2)))) =8.((Γ((3/2))Γ(2))/(Γ(2+(3/2)))){(1+Ψ(1)−{(2/5)+(2/3)+2+Ψ((1/2))})^2 +(ζ(2)−1((4/(25))+(4/9)+4+Ψ′((1/2))) Ψ(1)=−γ;Ψ((1/2))=−2ln(2)−γ ζ(2)=(π^2 /6) Ψ((1/2))=Σ_(n≥0) (1/(((1/2)+n)^2 ))=Σ_(n≥0) (4/((2n+1)^2 ))=4(1−(1/4))ζ(2) Γ(2+(3/2))=(1+(3/2))((3/2))Γ((3/2))...
$${x}={u}^{\mathrm{2}} \Rightarrow{dx}=\mathrm{2}{udu} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\sqrt{{x}}}{ln}^{\mathrm{2}} \left({x}\right){dx}.={A}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{2}{ln}\left({u}\right)\right)^{\mathrm{2}} .\mathrm{2}{udu} \\ $$$$=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} \left({u}\right).{udu} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} {u}^{{a}−\mathrm{1}} =\beta\left({a},\frac{\mathrm{3}}{\mathrm{2}}\right){du};{f}''\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} \left({u}\right).{u}^{{a}−\mathrm{1}} {du} \\ $$$${f}'\left({a}\right)=\beta\left({a},\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\Psi\left({a}\right)−\Psi\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$${f}''\left({a}\right)=\beta\left({a},\frac{\mathrm{3}}{\mathrm{2}}\right)\left\{\left(\Psi\left({a}\right)−\Psi\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)\right)+\left(\Psi'\left({a}\right)−\Psi'\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)\right\}\right. \\ $$$${A}=\mathrm{8}{f}''\left(\mathrm{2}\right) \\ $$$$=\mathrm{8}\left(\beta\left(\mathrm{2},\frac{\mathrm{3}}{\mathrm{2}}\right)\right)\left\{\left(\Psi\left(\mathrm{2}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\right)\right)^{\mathrm{2}} +\left(\Psi'\left(\mathrm{2}\right)−\Psi'\left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}\right)\right)\right) \\ $$$$=\mathrm{8}.\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{2}\right)}{\Gamma\left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\left\{\left(\mathrm{1}+\Psi\left(\mathrm{1}\right)−\left\{\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{2}+\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}\right)^{\mathrm{2}} +\left(\zeta\left(\mathrm{2}\right)−\mathrm{1}\left(\frac{\mathrm{4}}{\mathrm{25}}+\frac{\mathrm{4}}{\mathrm{9}}+\mathrm{4}+\Psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\right.\right. \\ $$$$\Psi\left(\mathrm{1}\right)=−\gamma;\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}{ln}\left(\mathrm{2}\right)−\gamma \\ $$$$\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{4}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\zeta\left(\mathrm{2}\right) \\ $$$$\Gamma\left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)… \\ $$$$ \\ $$$$ \\ $$

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