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Question Number 165152 by mnjuly1970 last updated on 26/Jan/22

prove Ω=∫_0 ^( 1) (( x − x^( 2) )/((1+x )ln(x))) dx = ln((4/π) ) −−−−−

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove} \\ $$$$\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}\:−\:{x}^{\:\mathrm{2}} }{\left(\mathrm{1}+{x}\:\right){ln}\left({x}\right)}\:{dx}\:=\:{ln}\left(\frac{\mathrm{4}}{\pi}\:\right) \\ $$$$\:\:\:−−−−− \\ $$

Answered by mindispower last updated on 26/Jan/22

f(a)=∫_0 ^1 ((x−x^(a+1) )/((1+x)ln(x)))dx,f(0)=0 Ω must bee <0 ln((4/π))>0 f′(a)=∫_0 ^1 ((−x^(1+a) )/(1+x)) =∫_0 ^1 ((x^(2+a) −x^(1+a) )/(1−x^2 ))dx =(1/2)∫_0 ^1 ((t^((a+1)/2) −t^(a/2) )/(1−t)) .dt recall Ψ(z+1)=−γ+∫_0 ^1 ((1−x^z )/(1−x))dx f^′ (a)=(1/2){Ψ((a/2)+1)−Ψ(((a+3)/2))} f(a)=ln(((Γ(((a+2)/2)))/(Γ(((a+3)/2)))))+c f(0)=0⇒c=−ln((2/( (√π))))=ln(((√π)/2)) Ω=f(1)=ln((((√π)/2)/1))=ln(((√π)/2))+ln(((√π)/2))=ln((π/4)) Ω=ln((π/4))

$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{x}^{{a}+\mathrm{1}} }{\left(\mathrm{1}+{x}\right){ln}\left({x}\right)}{dx},{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\Omega\:{must}\:{bee}\:<\mathrm{0}\:{ln}\left(\frac{\mathrm{4}}{\pi}\right)>\mathrm{0} \\ $$$${f}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}^{\mathrm{1}+{a}} }{\mathrm{1}+{x}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}+{a}} −{x}^{\mathrm{1}+{a}} }{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} −{t}^{\frac{{a}}{\mathrm{2}}} }{\mathrm{1}−{t}}\:.{dt} \\ $$$${recall}\:\Psi\left({z}+\mathrm{1}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{z}} }{\mathrm{1}−{x}}{dx} \\ $$$${f}^{'} \left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{\Psi\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)−\Psi\left(\frac{{a}+\mathrm{3}}{\mathrm{2}}\right)\right\} \\ $$$${f}\left({a}\right)={ln}\left(\frac{\Gamma\left(\frac{{a}+\mathrm{2}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}+\mathrm{3}}{\mathrm{2}}\right)}\right)+{c} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{c}=−{ln}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\right)={ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right) \\ $$$$\Omega={f}\left(\mathrm{1}\right)={ln}\left(\frac{\frac{\sqrt{\pi}}{\mathrm{2}}}{\mathrm{1}}\right)={ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)+{ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)={ln}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Omega={ln}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 27/Jan/22

thanks alot sir power ..grateful

$$\:\:\:\:{thanks}\:{alot}\:{sir}\:{power}\:..{grateful} \\ $$$$\:\: \\ $$

Commented by mindispower last updated on 27/Jan/22

wthe Pleasur Have a nice Day

$${wthe}\:{Pleasur}\:{Have}\:{a}\:{nice}\:{Day} \\ $$

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