Question Number 116667 by mnjuly1970 last updated on 05/Oct/20
$$\:\:\:\:\:\:\:\:\:\:\:\:...\:\:\:{nice}\:\:{calculus}... \\ $$$$ \\ $$$$\:\:\:\:\:\:{very}\:{nice}\:\:{integral}:: \\ $$$$\:\:\:\:\:\:\:{demonstrate}::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right){log}\left({x}\right)}\:{dx}\overset{???} {=}{log}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right) \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}.\mathrm{1970}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Answered by maths mind last updated on 06/Oct/20
![Hello after many tries Ω=∫_0 ^1 ((1−x)/((1+x+x^2 +x^3 ))).(dx/(log(x))) let f(a)=∫_0 ^1 ((1−x)/((1+x+x^2 +x^3 ))).(x^a /(log(x)))dx,a∈[0,1] Ω=f(0) f(a)=∫_0 ^1 (((1−x)^2 )/(1−x^4 ))..(x^a /(log(x)))dx f′(a)=∫_0 ^1 (((1−x)^2 )/((1−x^4 ))).∂_a x^a .(dx/(log(x)))=∫_0 ^1 (((x^2 −2x+1))/(1−x^4 )).x^a dx x^4 =t⇒dx=(t^(−(3/4)) /4)dt f′(a)=∫_0 ^1 ((t^(2/4) −2t^(1/4) +1)/(1−t))t^(a/4) .t^(−(3/4)) (dt/4) =−∫_0 ^1 ((1−t^((a/4)−(1/4)) )/(1−t))dt+2∫((1−t^((a/4)−(2/4)) )/(1−t))dt−∫((1−t^((a/4)−(3/4)) )/(1−t))dt we Have one of definition of Digamma Ψ(s+1)=−γ+∫_0 ^1 ((1−x^s )/(1−x))dx =(lnΓ(s+1))′ 4f′(a)=−Ψ(((a+3)/4))+2Ψ(((a+2)/4))−Ψ(((a+1)/4)) 4f(a)=−4lnΓ(((a+3)/4))−4lnΓ(((a+1)/4))+8lnΓ(((a+2)/4))+c lim_(a→∞) f(a)=0⇒c=0 f(0)=Ω 4f(0)=4Ω=−4lnΓ((3/4))−4lnΓ((1/4))+8lnΓ((1/2)) =−4ln(((Γ((1/4))Γ((3/4)))/(Γ^2 ((1/2))))):Γ((1/4))Γ(1−(1/4))=(π/(sin((π/4))))=π(√2) Γ((1/2))=(√π)⇒4Ω=−4ln(((π(√2))/(((√π))^2 )))=−4ln((√2))=4ln((√(1/2))) Ω=ln((√(1/2))) ∫_0 ^1 ((1−x)/(1+x+x^2 +x^3 )).(dx/(log(x)))=ln((√(1/2)))](Q116794.png)
$${Hello}\:{after}\:{many}\:{tries} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)}.\frac{{dx}}{{log}\left({x}\right)} \\ $$$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)}.\frac{{x}^{{a}} }{{log}\left({x}\right)}{dx},{a}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\Omega={f}\left(\mathrm{0}\right) \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }..\frac{{x}^{{a}} }{{log}\left({x}\right)}{dx} \\ $$$${f}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)}.\partial_{{a}} {x}^{{a}} .\frac{{dx}}{{log}\left({x}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }.{x}^{{a}} {dx} \\ $$$${x}^{\mathrm{4}} ={t}\Rightarrow{dx}=\frac{{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{4}}{dt} \\ $$$${f}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{2}}{\mathrm{4}}} −\mathrm{2}{t}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}}{\mathrm{1}−{t}}{t}^{\frac{{a}}{\mathrm{4}}} .{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} \frac{{dt}}{\mathrm{4}} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{\frac{{a}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{1}−{t}}{dt}+\mathrm{2}\int\frac{\mathrm{1}−{t}^{\frac{{a}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{4}}} }{\mathrm{1}−{t}}{dt}−\int\frac{\mathrm{1}−{t}^{\frac{{a}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{1}−{t}}{dt} \\ $$$${we}\:{Have}\:{one}\:{of}\:{definition}\:{of}\:{Digamma} \\ $$$$\Psi\left({s}+\mathrm{1}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{s}} }{\mathrm{1}−{x}}{dx}\:=\left({ln}\Gamma\left({s}+\mathrm{1}\right)\right)' \\ $$$$\mathrm{4}{f}'\left({a}\right)=−\Psi\left(\frac{{a}+\mathrm{3}}{\mathrm{4}}\right)+\mathrm{2}\Psi\left(\frac{{a}+\mathrm{2}}{\mathrm{4}}\right)−\Psi\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{4}{f}\left({a}\right)=−\mathrm{4}{ln}\Gamma\left(\frac{{a}+\mathrm{3}}{\mathrm{4}}\right)−\mathrm{4}{ln}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{4}}\right)+\mathrm{8}{ln}\Gamma\left(\frac{{a}+\mathrm{2}}{\mathrm{4}}\right)+{c} \\ $$$$\underset{{a}\rightarrow\infty} {\mathrm{lim}}{f}\left({a}\right)=\mathrm{0}\Rightarrow{c}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\Omega \\ $$$$\mathrm{4}{f}\left(\mathrm{0}\right)=\mathrm{4}\Omega=−\mathrm{4}{ln}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\mathrm{4}{ln}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{8}{ln}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\mathrm{4}{ln}\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\right):\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\pi\sqrt{\mathrm{2}} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\Rightarrow\mathrm{4}\Omega=−\mathrm{4}{ln}\left(\frac{\pi\sqrt{\mathrm{2}}}{\left(\sqrt{\pi}\right)^{\mathrm{2}} }\right)=−\mathrm{4}{ln}\left(\sqrt{\mathrm{2}}\right)=\mathrm{4}{ln}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right) \\ $$$$\Omega={ln}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }.\frac{{dx}}{{log}\left({x}\right)}={ln}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by maths mind last updated on 06/Oct/20
$${sam}\:{idea}\:{worck}\:{for}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{s}} }{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{k}} }.\frac{{dx}}{{log}\left({x}\right)} \\ $$
Commented by mnjuly1970 last updated on 07/Oct/20
$${thank}\:\:{you}\: \\ $$$${good}\:\:{for}\:{you} \\ $$$${your}\:{effort}\:\:{is}\:{admirable}... \\ $$