Menu Close

calculate-0-lnx-x-2-x-1-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 127236 by Bird last updated on 28/Dec/20
calculate ∫_0 ^∞ ((lnx)/((x^2 −x+1)^2 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{\left({x}^{\mathrm{2}} \:−{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Answered by mindispower last updated on 28/Dec/20
=∫_0 ^∞ ((ln(x)(1+x)^2 )/((x^3 +1)^2 ))dx=∫_0 ^∞ ((x^2 ln(x))/((1+x^3 )^2 ))+2∫_0 ^∞ ((ln(x).xdx)/((1+x^3 )^2 ))+∫((ln(x)dx)/((1+x^3 )^2 )) =a+b+c a=0:x→(1/x)⇒a=−a c= x=tg^(2/3) (t)⇒dx=(2/3)(1+tg^2 (t))tg^(−(1/3)) (t)dt ⇔∫_0 ^(π/2) (2/3).((ln(tg(t)))/((1+tg^2 (t))^2 )).(2/3)(1+tg^2 (t)).tg^(−(1/3)) (t)dt =(4/9)∫_0 ^(π/2) ((ln(tg(t)))/(1+tg^2 (t))).tg^(−(1/3)) dt =(4/9)∫_0 ^(π/2) ((ln(sin(x))−ln(cos(x)))/(cos^(−(1/3)) (t)))sin^(−(1/3)) (t).cos^2 (t)dt =(4/9).∫_0 ^(π/2) ln(sin(t))cos^(7/3) (t)sin^(−(1/3)) (t)−(4/9)∫_0 ^(π/2) ln(cos(t))cos^(7/3) (t)sin^(−(1/3)) (t)dt =(1/9)(A−B) β(x,y)=2∫_0 ^(π/2) cos(t)^(2x−1) sin^(2y−1) (t)dt ∂_x β(x,y)=2.∫_0 ^(π/2) 2ln(cos(t))cos(t)^(2x−1) sin^(2y−1) (t)dt A=(1/9)∂_y β(x,y)_((x,y)=((5/3),(1/3))) B=(1/9)∂_x β(x,y)_((x,y)=((5/3),(1/3))) ∂_x β(x,y)=β(x,y)[Ψ(x)−Ψ(x+y)) ((β((5/3),(1/3)))/9)(Ψ((1/3))−Ψ(2)−Ψ((5/3))+Ψ(2)) =((Γ((1/3))Γ((5/3)))/(9Γ(2)))(Ψ((1/3))−Ψ((5/3))) =((Γ((1/3)).(2/3)Γ((2/3)))/(9Γ(2)))(Ψ((1/3))−(3/2)−Ψ((2/3))) =(4/(27)).(π/( (√3)))(−(3/2)+πcot(((2π)/3)))=((4π)/(27(√3)))(−(3/2)−(π/( (√3)))) =((−2)/(81))π(3(√3)+2π) b=(8/9)∫_0 ^(π/2) ln(tg(t))((sin^((2/3) ) (t))/(cos^(2/3) (t)))cos^2 (t).tg^(−(1/3)) (t) =(8/9)(∫_0 ^(π/2) ln(sin(t))cos^(5/3) (t)sin^(1/3) (t)dt−∫_0 ^(π/2) ln(cos(t))cos^(5/3) (t)sin^(1/3) (t)dt) =(2/9)(Ψ((2/3))−Ψ(2)−Ψ((4/3))−Ψ(2))β((2/3),(4/3)) =(2/9)(Ψ((2/3))−Ψ((4/3))).((Γ((2/3))Γ((4/3)))/(Γ(2))) =(2/9)(Ψ((2/3))−3−Ψ((1/3))).(1/3)(π/(sin((π/3)))) =((2π)/(27))(πcot((π/3))−3).(2/( (√3)))=((4π)/(27(√3)))((π/( (√3)))−3).=((4π)/(81))(π−3(√3)) ∫_0 ^∞ ((ln(x))/((x^2 −x+1)^2 ))=((4π)/(81))(π−3(√3))−((2π)/(81))(3(√3)+2π) =((−18π(√3))/(81))=((−2π(√3))/9)≈−1.2092
$$=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} {ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} }+\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right).{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} }+\int\frac{{ln}\left({x}\right){dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} } \\ $$$$={a}+{b}+{c} \\ $$$${a}=\mathrm{0}:{x}\rightarrow\frac{\mathrm{1}}{{x}}\Rightarrow{a}=−{a} \\ $$$${c}= \\ $$$${x}={tg}^{\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)\Rightarrow{dx}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right){tg}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right){dt} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}}{\mathrm{3}}.\frac{{ln}\left({tg}\left({t}\right)\right)}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right)^{\mathrm{2}} }.\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right).{tg}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right){dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({tg}\left({t}\right)\right)}{\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)}.{tg}^{−\frac{\mathrm{1}}{\mathrm{3}}} {dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({sin}\left({x}\right)\right)−{ln}\left({cos}\left({x}\right)\right)}{{cos}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right)}{sin}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right).{cos}^{\mathrm{2}} \left({t}\right){dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}.\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({t}\right)\right){cos}^{\frac{\mathrm{7}}{\mathrm{3}}} \left({t}\right){sin}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right)−\frac{\mathrm{4}}{\mathrm{9}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\left({t}\right)\right){cos}^{\frac{\mathrm{7}}{\mathrm{3}}} \left({t}\right){sin}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left({A}−{B}\right) \\ $$$$\beta\left({x},{y}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left({t}\right)^{\mathrm{2}{x}−\mathrm{1}} {sin}^{\mathrm{2}{y}−\mathrm{1}} \left({t}\right){dt} \\ $$$$\partial_{{x}} \beta\left({x},{y}\right)=\mathrm{2}.\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{ln}\left({cos}\left({t}\right)\right){cos}\left({t}\right)^{\mathrm{2}{x}−\mathrm{1}} {sin}^{\mathrm{2}{y}−\mathrm{1}} \left({t}\right){dt} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{9}}\partial_{{y}} \beta\left({x},{y}\right)_{\left({x},{y}\right)=\left(\frac{\mathrm{5}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{9}}\partial_{{x}} \beta\left({x},{y}\right)_{\left({x},{y}\right)=\left(\frac{\mathrm{5}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\partial_{{x}} \beta\left({x},{y}\right)=\beta\left({x},{y}\right)\left[\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right) \\ $$$$\frac{\beta\left(\frac{\mathrm{5}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{9}}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\Psi\left(\mathrm{2}\right)−\Psi\left(\frac{\mathrm{5}}{\mathrm{3}}\right)+\Psi\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{3}}\right)}{\mathrm{9}\Gamma\left(\mathrm{2}\right)}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\Psi\left(\frac{\mathrm{5}}{\mathrm{3}}\right)\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right).\frac{\mathrm{2}}{\mathrm{3}}\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{9}\Gamma\left(\mathrm{2}\right)}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{3}}{\mathrm{2}}−\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right) \\ $$$$=\frac{\mathrm{4}}{\mathrm{27}}.\frac{\pi}{\:\sqrt{\mathrm{3}}}\left(−\frac{\mathrm{3}}{\mathrm{2}}+\pi{cot}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)=\frac{\mathrm{4}\pi}{\mathrm{27}\sqrt{\mathrm{3}}}\left(−\frac{\mathrm{3}}{\mathrm{2}}−\frac{\pi}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=\frac{−\mathrm{2}}{\mathrm{81}}\pi\left(\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{2}\pi\right) \\ $$$${b}=\frac{\mathrm{8}}{\mathrm{9}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({tg}\left({t}\right)\right)\frac{{sin}^{\frac{\mathrm{2}}{\mathrm{3}}\:} \left({t}\right)}{{cos}^{\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)}{cos}^{\mathrm{2}} \left({t}\right).{tg}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right) \\ $$$$=\frac{\mathrm{8}}{\mathrm{9}}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({t}\right)\right){cos}^{\frac{\mathrm{5}}{\mathrm{3}}} \left({t}\right){sin}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right){dt}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\left({t}\right)\right){cos}^{\frac{\mathrm{5}}{\mathrm{3}}} \left({t}\right){sin}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({t}\right){dt}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{9}}\left(\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\Psi\left(\mathrm{2}\right)−\Psi\left(\frac{\mathrm{4}}{\mathrm{3}}\right)−\Psi\left(\mathrm{2}\right)\right)\beta\left(\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{9}}\left(\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\Psi\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\right).\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{3}}\right)}{\Gamma\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{9}}\left(\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\mathrm{3}−\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right).\frac{\mathrm{1}}{\mathrm{3}}\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{27}}\left(\pi{cot}\left(\frac{\pi}{\mathrm{3}}\right)−\mathrm{3}\right).\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{4}\pi}{\mathrm{27}\sqrt{\mathrm{3}}}\left(\frac{\pi}{\:\sqrt{\mathrm{3}}}−\mathrm{3}\right).=\frac{\mathrm{4}\pi}{\mathrm{81}}\left(\pi−\mathrm{3}\sqrt{\mathrm{3}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{4}\pi}{\mathrm{81}}\left(\pi−\mathrm{3}\sqrt{\mathrm{3}}\right)−\frac{\mathrm{2}\pi}{\mathrm{81}}\left(\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{2}\pi\right) \\ $$$$=\frac{−\mathrm{18}\pi\sqrt{\mathrm{3}}}{\mathrm{81}}=\frac{−\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\approx−\mathrm{1}.\mathrm{2092} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Dec/20
thank you sir for this hardwork
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{this}\:\mathrm{hardwork} \\ $$
Commented by mindispower last updated on 29/Dec/20
withe pleasur
$${withe}\:{pleasur} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *