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Advanced-Calculus-evaluation-the-value-of-0-pi-2-sin-2-x-ln-sin-x-dx-solution-a-0

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Question Number 140588 by mnjuly1970 last updated on 09/May/21
.......Advanced ....β˜…β˜…β˜…....Calculus....... evaluation the value of : 𝛗 :=∫_0 ^( (Ο€/2)) sin^2 (x).ln(sin(x))dx solution:: ΞΎ (a):=∫_0 ^( (Ο€/2)) sin^(2+a) (x)dx =(1/2)Ξ² (((3+a)/2) ,(1/2)) :=(1/2)(((Ξ“(((3+a)/2))Ξ“((1/2)))/(Ξ“(2+(a/2))))).......βœ“ 𝛗:= ΞΎ β€² (0) ..............βœ“ :=(1/2) (βˆšΟ€) (( Ξ“β€²(((3+a)/2)).Ξ“(2+(a/2))βˆ’Ξ“(((3+a)/2)).Ξ“β€²(2+(a/2)))/(Ξ“^2 (2+(a/2)))) ∣_(a=0) :=(1/2)(βˆšΟ€) ((Ξ“β€²((3/2))βˆ’Ξ“((3/2)).Ξ“β€²(2))/(( Ξ“^2 (2):=1 ))) :=(1/2)(βˆšΟ€) ((ψ((3/2))Ξ“((3/2))βˆ’Ξ“((3/2)).ψ(2))/1) := ((βˆšΟ€)/4){ (2βˆ’Ξ³βˆ’2ln(2)βˆ’(1βˆ’Ξ³)} :=((βˆšΟ€)/4)(1βˆ’ln(4))=(βˆšΟ€) ln(((e/4))^(1/4) )
$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….{Advanced}\:….\bigstar\bigstar\bigstar….{Calculus}……. \\ $$$$\:\:\:\:\:\:\:\:{evaluation}\:{the}\:{value}\:{of}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\::=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \left({x}\right).{ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\:\:\xi\:\left({a}\right):=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}+{a}} \left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\beta\:\left(\frac{\mathrm{3}+{a}}{\mathrm{2}}\:,\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\Gamma\left(\frac{\mathrm{3}+{a}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)}\right)…….\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\:\xi\:'\:\left(\mathrm{0}\right)\:…………..\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\pi}\:\frac{\:\Gamma'\left(\frac{\mathrm{3}+{a}}{\mathrm{2}}\right).\Gamma\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)βˆ’\Gamma\left(\frac{\mathrm{3}+{a}}{\mathrm{2}}\right).\Gamma'\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)}\:\mid_{{a}=\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\:\:\frac{\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)βˆ’\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right).\Gamma'\left(\mathrm{2}\right)}{\left(\:\Gamma^{\mathrm{2}} \left(\mathrm{2}\right):=\mathrm{1}\:\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\:\frac{\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)βˆ’\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right).\psi\left(\mathrm{2}\right)}{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\::=\:\frac{\sqrt{\pi}}{\mathrm{4}}\left\{\:\left(\mathrm{2}βˆ’\gammaβˆ’\mathrm{2}{ln}\left(\mathrm{2}\right)βˆ’\left(\mathrm{1}βˆ’\gamma\right)\right\}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\::=\frac{\sqrt{\pi}}{\mathrm{4}}\left(\mathrm{1}βˆ’{ln}\left(\mathrm{4}\right)\right)=\sqrt{\pi}\:{ln}\left(\sqrt[{\mathrm{4}}]{\frac{{e}}{\mathrm{4}}}\right) \\ $$
Answered by mathmax by abdo last updated on 10/May/21
Ξ¦=∫_0 ^(Ο€/2) sin^2 x log(sinx)dx let f(a)=∫_0 ^(Ο€/2) sin^a x dx β‡’ f(a)=∫_0 ^(Ο€/2) e^(alog(sinx)) dx β‡’f^β€² (a)=∫_0 ^(Ο€/2) log(sinx) sin^a x dx β‡’ f^β€² (2)=∫_0 ^(Ο€/2) sin^2 x log(sinx)dx =Ξ¦ we have f(a)=∫_0 ^(Ο€/2) sin^(2(((a+1)/2))βˆ’1) cos^(2((1/2))βˆ’1) x dx =(1/2)B(((a+1)/2),(1/2)) =(1/2).((Ξ“(((a+1)/2)).Ξ“((1/2)))/(Ξ“(((a+1)/(2 ))+(1/2))))=((βˆšΟ€)/2).((Ξ“(((a+1)/2)))/(Ξ“((a/2)+1))) =((βˆšΟ€)/2)Γ—((Ξ“(((a+1)/2)))/((a/2)Ξ“((a/2)))) =((βˆšΟ€)/a) Γ—((Ξ“(((a+1)/2)))/(Ξ“((a/2)))) β‡’ f^β€² (a) =βˆ’((βˆšΟ€)/a^2 )Γ—((Ξ“(((a+1)/2)))/(Ξ“((a/2))))+((βˆšΟ€)/a)Γ—(((1/2)Ξ“^β€² (((a+1)/2)).Ξ“((a/2))βˆ’(1/2)Ξ“^β€² ((a/2))Ξ“(((a+1)/2)))/(Ξ“^2 ((a/2)))) β‡’ Ξ¦=f^β€² (2)=βˆ’((βˆšΟ€)/4)Γ—((Ξ“((3/2)))/1) +((βˆšΟ€)/4)Γ—((Ξ“^β€² ((3/2)).1βˆ’Ξ“^β€² (1).Ξ“((3/2)))/1^2 ) =βˆ’((βˆšΟ€)/4).Ξ“((3/2)) +((βˆšΟ€)/4)Γ—(Ξ“^β€² ((3/2))+Ξ³ .Ξ“((3/2))) Ξ“^β€² (1)=∫_0 ^∞ e^(βˆ’t) logt dt =βˆ’Ξ³
$$\Phi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{log}\left(\mathrm{sinx}\right)\mathrm{dx}\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{a}} \mathrm{x}\:\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{\mathrm{alog}\left(\mathrm{sinx}\right)} \:\mathrm{dx}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{log}\left(\mathrm{sinx}\right)\:\mathrm{sin}^{\mathrm{a}} \mathrm{x}\:\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{2}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{log}\left(\mathrm{sinx}\right)\mathrm{dx}\:=\Phi \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\mathrm{sin}^{\mathrm{2}\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)βˆ’\mathrm{1}} \:\mathrm{cos}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)βˆ’\mathrm{1}} \mathrm{x}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{B}\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}\:}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{a}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}Γ—\frac{\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)}{\frac{\mathrm{a}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{a}}{\mathrm{2}}\right)}\:=\frac{\sqrt{\pi}}{\mathrm{a}}\:Γ—\frac{\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{a}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=βˆ’\frac{\sqrt{\pi}}{\mathrm{a}^{\mathrm{2}} }Γ—\frac{\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{a}}{\mathrm{2}}\right)}+\frac{\sqrt{\pi}}{\mathrm{a}}Γ—\frac{\frac{\mathrm{1}}{\mathrm{2}}\Gamma^{'} \left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right).\Gamma\left(\frac{\mathrm{a}}{\mathrm{2}}\right)βˆ’\frac{\mathrm{1}}{\mathrm{2}}\Gamma^{'} \left(\frac{\mathrm{a}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{a}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\:\Phi=\mathrm{f}^{'} \left(\mathrm{2}\right)=βˆ’\frac{\sqrt{\pi}}{\mathrm{4}}Γ—\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{1}}\:+\frac{\sqrt{\pi}}{\mathrm{4}}Γ—\frac{\Gamma^{'} \left(\frac{\mathrm{3}}{\mathrm{2}}\right).\mathrm{1}βˆ’\Gamma^{'} \left(\mathrm{1}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{1}^{\mathrm{2}} } \\ $$$$=βˆ’\frac{\sqrt{\pi}}{\mathrm{4}}.\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:+\frac{\sqrt{\pi}}{\mathrm{4}}Γ—\left(\Gamma^{'} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\gamma\:.\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$\Gamma^{'} \left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{βˆ’\mathrm{t}} \:\mathrm{logt}\:\mathrm{dt}\:=βˆ’\gamma \\ $$

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