Question Number 28889 by abdo imad last updated on 31/Jan/18
$${find}\:\:{I}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left({x}−{e}^{{i}\theta} \right){d}\theta\:\:\:\:{and}\:{xfromR}\:{and}\:{x}^{\mathrm{2}} \neq\mathrm{1}. \\ $$
Commented by abdo imad last updated on 02/Feb/18
![let put f(x)= ∫_0 ^(2π) ln(x −e^(iθ) )dθ we have f^′ (x)= ∫_0 ^(2π) (dθ/( x−e^(iθ) ))= ∫_0 ^(2π) (dθ/(x −cosθ −isinθ)) = ∫_0 ^(2π) ((x−cosθ +i sinθ)/((x−cosθ)^2 +sin^2 θ))dθ =∫_0 ^(2π) ((x−cosθ)/(x^2 −2xcosθ +1))dθ +i∫_0 ^(2π) ((sinθ)/(x^2 −2xcosθ +1))dθ =A(x) +iB(x) A(x)=(1/2) ∫_0 ^(2π) ((2x−2cosθ)/(x^2 −2xcosθ+1))dθ=(1/2)[ln∣x^2 −2xcosθ+1∣]_(θ=0) ^(2π) =0 let find B(x) the ch. e^(iθ) =z give B(x)=∫_(∣z∣=1) (((z−z^(−1) )/(2i))/(x^2 −2x((z+z^(−1) )/2) +1))(dz/(iz)) = ∫_(∣z∣=1) ((z−z^(−1) )/(2i( x^2 −2x((z+z^(−1) )/2)+1)iz))dz =∫_(∣z∣=1) ((z^(−1) −z)/(z(2x^2 −2xz −2xz^(−1) +2)))dz = ∫_(∣z∣=1) (((1/z)−z)/(2(x^2 z−xz^2 −x +z)))dz =∫_(∣z∣=1) ((1−z^2 )/(2z( −xz^2 +(1+x^2 )z −x)))dz =∫_(∣z∣=1) ((z^2 −1)/(2z( xz^2 −(1+x^2 )z +x)))dz let introduce the complex function ψ(z)= ((z^2 −1)/(2z( xz^2 −(1+x^2 )z +x))) poles ofψ? xz^2 −(1+x^2 )z +x=0⇒Δ=(1+x^2 )^2 −4x^2 =1+2x^2 +x^4 −4x^2 =(1−x^2 )^2 ⇒z_(1 ) =((1+x^2 +∣1−x^2 ∣)/(2x)) and z_2 =((1+x^2 −∣1−x^2 ∣)/(2x)) (x≠0) case 1 if ∣x∣<1 ⇒z_1 = (1/x) and z_2 =x and we have∣z_1 ∣>1and ∣z_2 ∣<1 ψ(z)= ((z^2 −1)/(2xz(z−z_1 )(z−z_2 ))) and by residus theorem ∫_(∣z∣=1) ψ(z)dz=2iπ(Res(ψ,0) +Res(ψ,z_2 )) Res(ψ,0)=lim_(z→0) zψ(z)= ((−1)/(2x z_1 .z_2 ))=((−1)/(2x)) Res(ψ,z_2 )=lim_(z→z_2 ) (z−z_2 )ψ(z)= ((z_2 ^2 −1)/(2xz_2 (z_2 −z_1 ))) = ((x^2 −1)/(2x^2 (x−(1/x))))=((x^2 −1)/(2x(x^2 −1)))= (1/(2x)) ∫_(∣z∣=1) ψ(z)dz=2iπ(0)=0 case 2 if ∣x∣>1⇒ z_1 = x and z_2 =(1/x) and ∣z_1 ∣>1 and ∣z_2 ∣<1 ∫_(∣z∣=1) ψ(z)dz=2iπ( Res(ψ,0)+Res(ψ,z_2 )) Res(ψ,0)= ((−1)/(2x )) and Res(ψ,z_2 )=((z_2 ^2 −1)/(2xz_2 (z_2 −z_1 ))) =(((1/x^2 )−1)/(2((1/x)−x)))= ((1−x^2 )/(2x^2 ((1−x^2 )/x))) = (1/(2x)) and ∫_(∣z∣=1) ψ(z)dz=0so A(x)=B(x)=0⇒f^′ (x)=0 ⇒f(x)=λ ∀ x /x^2 ≠1](https://www.tinkutara.com/question/Q28945.png)
$${let}\:{put}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{ln}\left({x}\:−{e}^{{i}\theta} \right){d}\theta\:{we}\:{have} \\ $$$${f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{d}\theta}{\:{x}−{e}^{{i}\theta} }=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{d}\theta}{{x}\:−{cos}\theta\:−{isin}\theta} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{x}−{cos}\theta\:+{i}\:{sin}\theta}{\left({x}−{cos}\theta\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{x}−{cos}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}}{d}\theta\:+{i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{sin}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}}{d}\theta \\ $$$$={A}\left({x}\right)\:+{iB}\left({x}\right) \\ $$$${A}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{2}{x}−\mathrm{2}{cos}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta+\mathrm{1}}{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta+\mathrm{1}\mid\right]_{\theta=\mathrm{0}} ^{\mathrm{2}\pi} \\ $$$$=\mathrm{0}\:\:{let}\:{find}\:{B}\left({x}\right)\:{the}\:{ch}.\:{e}^{{i}\theta} ={z}\:{give} \\ $$$${B}\left({x}\right)=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\:+\mathrm{1}}\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}\left(\:{x}^{\mathrm{2}} −\mathrm{2}{x}\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}+\mathrm{1}\right){iz}}{dz} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{z}^{−\mathrm{1}} −{z}}{{z}\left(\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{2}{xz}\:−\mathrm{2}{xz}^{−\mathrm{1}} +\mathrm{2}\right)}{dz} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\frac{\mathrm{1}}{{z}}−{z}}{\mathrm{2}\left({x}^{\mathrm{2}} {z}−{xz}^{\mathrm{2}} −{x}\:+{z}\right)}{dz} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{2}{z}\left(\:−{xz}^{\mathrm{2}} \:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:−{x}\right)}{dz} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{z}^{\mathrm{2}} \:−\mathrm{1}}{\mathrm{2}{z}\left(\:{xz}^{\mathrm{2}} \:−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:+{x}\right)}{dz}\:{let}\:{introduce}\:{the} \\ $$$${complex}\:{function}\:\psi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{z}\left(\:{xz}^{\mathrm{2}} −\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:+{x}\right)}\:\:{poles}\:{of}\psi? \\ $$$${xz}^{\mathrm{2}} \:−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:+{x}=\mathrm{0}\Rightarrow\Delta=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} \\ $$$$=\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:−\mathrm{4}{x}^{\mathrm{2}} \:=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \Rightarrow{z}_{\mathrm{1}\:} =\frac{\mathrm{1}+{x}^{\mathrm{2}} \:+\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}{\mathrm{2}{x}}\:{and} \\ $$$${z}_{\mathrm{2}} =\frac{\mathrm{1}+{x}^{\mathrm{2}} −\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}{\mathrm{2}{x}}\:\:\:\:\left({x}\neq\mathrm{0}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${case}\:\mathrm{1}\:{if}\:\:\mid{x}\mid<\mathrm{1}\:\:\:\:\:\Rightarrow{z}_{\mathrm{1}} =\:\frac{\mathrm{1}}{{x}}\:{and}\:{z}_{\mathrm{2}} ={x}\:{and}\:{we}\:{have}\mid{z}_{\mathrm{1}} \mid>\mathrm{1}{and}\:\mid{z}_{\mathrm{2}} \mid<\mathrm{1} \\ $$$$\psi\left({z}\right)=\:\:\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{xz}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:{and}\:{by}\:{residus}\:{theorem} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \psi\left({z}\right){dz}=\mathrm{2}{i}\pi\left({Res}\left(\psi,\mathrm{0}\right)\:+{Res}\left(\psi,{z}_{\mathrm{2}} \right)\right) \\ $$$${Res}\left(\psi,\mathrm{0}\right)={lim}_{{z}\rightarrow\mathrm{0}} {z}\psi\left({z}\right)=\:\frac{−\mathrm{1}}{\mathrm{2}{x}\:{z}_{\mathrm{1}} .{z}_{\mathrm{2}} }=\frac{−\mathrm{1}}{\mathrm{2}{x}} \\ $$$${Res}\left(\psi,{z}_{\mathrm{2}} \right)={lim}_{{z}\rightarrow{z}_{\mathrm{2}} } \:\left({z}−{z}_{\mathrm{2}} \right)\psi\left({z}\right)=\:\frac{{z}_{\mathrm{2}} ^{\mathrm{2}} \:−\mathrm{1}}{\mathrm{2}{xz}_{\mathrm{2}} \left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)} \\ $$$$=\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} \left({x}−\frac{\mathrm{1}}{{x}}\right)}=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}=\:\frac{\mathrm{1}}{\mathrm{2}{x}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \psi\left({z}\right){dz}=\mathrm{2}{i}\pi\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${case}\:\mathrm{2}\:\:{if}\:\mid{x}\mid>\mathrm{1}\Rightarrow\:{z}_{\mathrm{1}} =\:{x}\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{1}}{{x}}\:{and}\:\mid{z}_{\mathrm{1}} \mid>\mathrm{1}\:{and}\:\mid{z}_{\mathrm{2}} \mid<\mathrm{1} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \psi\left({z}\right){dz}=\mathrm{2}{i}\pi\left(\:{Res}\left(\psi,\mathrm{0}\right)+{Res}\left(\psi,{z}_{\mathrm{2}} \right)\right) \\ $$$${Res}\left(\psi,\mathrm{0}\right)=\:\frac{−\mathrm{1}}{\mathrm{2}{x}\:}\:{and}\:{Res}\left(\psi,{z}_{\mathrm{2}} \right)=\frac{{z}_{\mathrm{2}} ^{\mathrm{2}} \:−\mathrm{1}}{\mathrm{2}{xz}_{\mathrm{2}} \left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)} \\ $$$$=\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{1}}{{x}}−{x}\right)}=\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{x}}}\:=\:\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:{and}\:\int_{\mid{z}\mid=\mathrm{1}} \psi\left({z}\right){dz}=\mathrm{0}{so} \\ $$$${A}\left({x}\right)={B}\left({x}\right)=\mathrm{0}\Rightarrow{f}^{'} \left({x}\right)=\mathrm{0}\:\Rightarrow{f}\left({x}\right)=\lambda\:\:\forall\:{x}\:/{x}^{\mathrm{2}} \neq\mathrm{1} \\ $$
Commented by abdo imad last updated on 02/Feb/18
![λ=f(0)=∫_0 ^(2π) ln(−e^(iθ) )dθ = ∫_0 ^(2π) ln(e^(i(π+θ)) )dθ =∫_0 ^(2π) i(π+θ)dθ=2iπ^2 +i[(θ^2 /2)]_0 ^(2π) =2iπ^2 +2iπ^2 =4iπ^2 .](https://www.tinkutara.com/question/Q28946.png)
$$\lambda={f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left(−{e}^{{i}\theta} \right){d}\theta\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{ln}\left({e}^{{i}\left(\pi+\theta\right)} \right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {i}\left(\pi+\theta\right){d}\theta=\mathrm{2}{i}\pi^{\mathrm{2}} \:+{i}\left[\frac{\theta^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}\pi} =\mathrm{2}{i}\pi^{\mathrm{2}} \:+\mathrm{2}{i}\pi^{\mathrm{2}} =\mathrm{4}{i}\pi^{\mathrm{2}} . \\ $$
Commented by abdo imad last updated on 02/Feb/18
$${if}\:{z}={r}\:{e}^{{i}\theta} \:\:\:\:\:\:\:{ln}\left({z}\right)={ln}\left({r}\right)\:+{i}\theta\:\:\:\:\:\:\:\:\:\:\:−\pi<\theta<\pi\:.{and}\:{r}>\mathrm{0} \\ $$