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Question-172359

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Question Number 172359 by mnjuly1970 last updated on 25/Jun/22
Answered by Mathspace last updated on 26/Jun/22
J=∫_0 ^1 x^(−(1/2)) (1−x)^(−(1/2)) ln^2 xdx let f(a)=∫_0 ^1 x^(a−(1/2)) (1−x)^(−(1/2)) dx we have f^(′′) (a)=∫_0 ^1 x^(a−(1/2)) (1−x)^(−(1/2)) ln^2 xdx ⇒f^(′′) (0)=∫_0 ^1 x^(−(1/2)) (1−x)^(−(1/2)) ln^2 x dx f(a)=∫_0 ^1 x^(a+(1/2)−1) (1−x)^((1/2)−1) dx =B(a+(1/2),(1/2))=((Γ(a+(1/2))Γ((1/2)))/(Γ(a+1))) =(√π)×((Γ(a+(1/2)))/(Γ(a+1))) ⇒f^′ (a)=(√π)×((Γ^′ (a+(1/2))Γ(a+1)−Γ(a+(1/2))Γ^′ (a+1))/(Γ^2 (a+1))) we/have ψ(x)=((Γ^′ (x))/(Γ(x))) ⇒ f^′ (a)=(√π)×((ψ(a+(1/2))Γ(a+(1/2))Γ(a+1)−Γ(a+(1/2))ψ(a+1)Γ(a+1))/(Γ^2 (a+1)))
$${J}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} {xdx} \\ $$$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$${we}\:{have}\:{f}^{''} \left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} {xdx} \\ $$$$\Rightarrow{f}^{''} \left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} {x}\:{dx} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx} \\ $$$$={B}\left({a}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({a}+\mathrm{1}\right)} \\ $$$$=\sqrt{\pi}×\frac{\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({a}+\mathrm{1}\right)} \\ $$$$\Rightarrow{f}^{'} \left({a}\right)=\sqrt{\pi}×\frac{\Gamma^{'} \left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}+\mathrm{1}\right)−\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma^{'} \left({a}+\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left({a}+\mathrm{1}\right)} \\ $$$${we}/{have}\:\psi\left({x}\right)=\frac{\Gamma^{'} \left({x}\right)}{\Gamma\left({x}\right)}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=\sqrt{\pi}×\frac{\psi\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}+\mathrm{1}\right)−\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left({a}+\mathrm{1}\right)} \\ $$

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