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Question Number 62145 by maxmathsup by imad last updated on 16/Jun/19

calculate ∫_0 ^π ln(x^2 −2xsinθ +1)dθ

$${calculate}\:\:\int_{\mathrm{0}} ^{\pi} {ln}\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta\:+\mathrm{1}\right){d}\theta \\ $$

Commented by maxmathsup by imad last updated on 17/Jun/19

let f(x) =∫_0 ^π ln(x^2 −2xsinθ +1)dθ ⇒f^′ (x) =∫_0 ^π ((2x−2sinθ)/(x^2 −2xsinθ+1)) dθ ⇒ ((f^′ (x))/2) =∫_0 ^π ((x−sinθ)/(x^2 −2xsinθ +1))dθ =_(θ=(t/2)) ∫_0 ^(2π) ((x−sin((t/2)))/(x^2 −2x sin((t/2))+1)) (dt/2) ⇒ f^′ (x) =_(z =e^((it)/2) ) ∫_(∣z∣=1) ((x−((z−z^(−1) )/(2i)))/(x^2 −2x((z−z^(−1) )/(2i)))) ((2dz)/(iz)) = ∫_(∣z∣=1) ((2ix−z+z^(−1) )/(−z{ 2ix^2 −2xz +2xz^(−1) })) (2i)dz =−∫_(∣z∣=1) ((−4x−2iz +2iz^(−1) )/(2ix^2 z−2xz^2 +2x))dz =∫ ((4x+2iz−2iz^(−1) )/(−2xz^2 +2ix^2 z +2x))dz =(1/x)∫_(∣z∣=1) ((4xz +2iz^2 −2i)/(z(−2xz^2 +2ix^2 z +2x}))dz (x≠0) =−(1/x)∫_(∣z∣=1) ((iz^2 +2xz −i)/(z{z^2 −ixz −1}))dz let w(z)=((iz^2 +2xz −i)/(z{z^2 −ixz −1})) poles of w? z^2 −ixz −1 =0→Δ^′ =(−ix)^2 +1 =1−x^2 case 1 x>1 ⇒ Δ^′ =(i(√(x^2 −1)))^2 ⇒z_1 =ix+i(√(x^2 −1)) z_2 =ix−i(√(x^2 −1)) and w(z)=((iz^2 +2xz−i)/(z(z−z_1 )(z−z_2 ))) ∣z_1 ∣ −1=∣x+(√(x^2 −1))∣ =x+(√(x^2 −1))−1 =x−1 +(√(x^2 −1))>0 (z_1 is out of circle) ∣z_2 ∣−1 =∣x−(√(x^2 −1))∣−1 =x−1−(√(x^2 −1)) (x−1)^2 −(x^2 −1) =x^2 −2x +1−x^2 +1 =2−2x =2(1−x)<0 ⇒∣z_2 ∣<1 ∫_(∣z∣=1) w(z)dz =2iπ { Res(w,o) +Res(w,z_2 )} Res(w,0) =lim_(z→0) zw(z) =((−i)/(z_1 z_2 )) =((−i)/(−1)) =i Res(w,z_2 ) =lim_(z→z_2 ) (z−z_2 )w(z) =((iz_2 ^2 +2xz_2 −i)/(z_2 (z_2 −z_1 ))) =((iz_2 ^2 +2xz_2 −i)/((ix−i(√(x^2 −1)))^2 +1)) =((i(ix−i(√(x^2 −1)))^2 +2x(ix−(√(x^2 −1))) −i)/(1−(x+(√(x^2 −1)))^2 )) =((−i(x+(√(x^2 −1)))^2 +2ix^2 −2x(√(x^2 −1))−i)/(1−(x^2 +2x(√(x^2 −1))+x^2 −1))) =.... be continued... m

$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\pi} {ln}\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta\:+\mathrm{1}\right){d}\theta\:\Rightarrow{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{2}{x}−\mathrm{2}{sin}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta+\mathrm{1}}\:{d}\theta\:\Rightarrow \\ $$$$\frac{{f}^{'} \left({x}\right)}{\mathrm{2}}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}−{sin}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta\:+\mathrm{1}}{d}\theta\:=_{\theta=\frac{{t}}{\mathrm{2}}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{x}−{sin}\left(\frac{{t}}{\mathrm{2}}\right)}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:{sin}\left(\frac{{t}}{\mathrm{2}}\right)+\mathrm{1}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$\Rightarrow\:{f}^{'} \left({x}\right)\:=_{{z}\:={e}^{\frac{{it}}{\mathrm{2}}} } \:\:\:\:\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{x}−\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}\:\frac{\mathrm{2}{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{2}{ix}−{z}+{z}^{−\mathrm{1}} }{−{z}\left\{\:\mathrm{2}{ix}^{\mathrm{2}} −\mathrm{2}{xz}\:+\mathrm{2}{xz}^{−\mathrm{1}} \right\}}\:\left(\mathrm{2}{i}\right){dz} \\ $$$$=−\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−\mathrm{4}{x}−\mathrm{2}{iz}\:+\mathrm{2}{iz}^{−\mathrm{1}} }{\mathrm{2}{ix}^{\mathrm{2}} {z}−\mathrm{2}{xz}^{\mathrm{2}} \:+\mathrm{2}{x}}{dz} \\ $$$$=\int\:\frac{\mathrm{4}{x}+\mathrm{2}{iz}−\mathrm{2}{iz}^{−\mathrm{1}} }{−\mathrm{2}{xz}^{\mathrm{2}} \:+\mathrm{2}{ix}^{\mathrm{2}} {z}\:+\mathrm{2}{x}}{dz}\:=\frac{\mathrm{1}}{{x}}\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}{xz}\:+\mathrm{2}{iz}^{\mathrm{2}} \:−\mathrm{2}{i}}{{z}\left(−\mathrm{2}{xz}^{\mathrm{2}} \:+\mathrm{2}{ix}^{\mathrm{2}} {z}\:+\mathrm{2}{x}\right\}}{dz}\:\:\:\left({x}\neq\mathrm{0}\right) \\ $$$$=−\frac{\mathrm{1}}{{x}}\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{iz}^{\mathrm{2}} \:+\mathrm{2}{xz}\:−{i}}{{z}\left\{{z}^{\mathrm{2}} −{ixz}\:−\mathrm{1}\right\}}{dz}\:\:{let}\:{w}\left({z}\right)=\frac{{iz}^{\mathrm{2}} \:+\mathrm{2}{xz}\:−{i}}{{z}\left\{{z}^{\mathrm{2}} −{ixz}\:−\mathrm{1}\right\}}\:\:{poles}\:{of}\:{w}? \\ $$$${z}^{\mathrm{2}} −{ixz}\:−\mathrm{1}\:=\mathrm{0}\rightarrow\Delta^{'} \:=\left(−{ix}\right)^{\mathrm{2}} +\mathrm{1}\:=\mathrm{1}−{x}^{\mathrm{2}} \\ $$$${case}\:\mathrm{1}\:\:\:{x}>\mathrm{1}\:\Rightarrow\:\Delta^{'} \:=\left({i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} ={ix}+{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${z}_{\mathrm{2}} ={ix}−{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:{and}\:{w}\left({z}\right)=\frac{{iz}^{\mathrm{2}} +\mathrm{2}{xz}−{i}}{{z}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}=\mid{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid\:={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}\:={x}−\mathrm{1}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}>\mathrm{0}\:\left({z}_{\mathrm{1}} {is}\:{out}\:{of}\:{circle}\right) \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\mid{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid−\mathrm{1}\:={x}−\mathrm{1}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:={x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{1}\:=\mathrm{2}−\mathrm{2}{x}\:=\mathrm{2}\left(\mathrm{1}−{x}\right)<\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{2}} \mid<\mathrm{1} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({w},{o}\right)\:+{Res}\left({w},{z}_{\mathrm{2}} \right)\right\} \\ $$$${Res}\left({w},\mathrm{0}\right)\:={lim}_{{z}\rightarrow\mathrm{0}} {zw}\left({z}\right)\:=\frac{−{i}}{{z}_{\mathrm{1}} {z}_{\mathrm{2}} }\:=\frac{−{i}}{−\mathrm{1}}\:={i}\: \\ $$$${Res}\left({w},{z}_{\mathrm{2}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{2}} } \:\:\:\left({z}−{z}_{\mathrm{2}} \right){w}\left({z}\right)\:=\frac{{iz}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{2}{xz}_{\mathrm{2}} −{i}}{{z}_{\mathrm{2}} \left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)}\:=\frac{{iz}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{2}{xz}_{\mathrm{2}} −{i}}{\left({ix}−{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{{i}\left({ix}−{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{2}{x}\left({ix}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:−{i}}{\mathrm{1}−\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} } \\ $$$$=\frac{−{i}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{2}{ix}^{\mathrm{2}} \:−\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−{i}}{\mathrm{1}−\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+{x}^{\mathrm{2}} −\mathrm{1}\right)}\:=....\:{be}\:{continued}... \\ $$$$ \\ $$$${m} \\ $$

Answered by perlman last updated on 16/Jun/19

let A(θ)=∫_0 ^π ln(x^2 −2xcos(θ)+1)dθ θ∈IR\{2kπ,k∈IZ} A(θ)=[xln(x^2 −2cos(θ)x+1)]_0 ^π −∫((2x^2 −2xcos(θ))/(x^2 −2cos(θ)x+1))dx =πln(π^2 −2cos(θ)π+1)−2∫_0 ^π ((x^2 −xcos(θ))/(x^2 −2cos(θ)x+1))dx ((x^2 −xcos(θ))/(x^2 −2cos(θ)x+1))=1+((xcos(θ)−1)/(x^2 −2cos(θ)x+1)) A(θ)=πln(π^2 −2cos(θ)π+1)−2∫_0 ^π 1+((xcos(θ)−1)/(x^2 −2cos(θ)x+1))dx the 2 nd integral =−2π −2∫_0 ^π ((xcos(θ)−1)/(x^2 −2cos(θ)x+1))dx=−2π−∫_0 ^π ((cos(θ)[2x−2cos(θ)]−2+2cos^2 (θ))/(x^2 −2cos(θ)x+1))dx =−2π−cos(θ)∫_0 ^π ((2x−2cos(θ))/(x^2 −2cos(θ)x+1))dx+2∫((sin^2 (θ))/(x^2 −2cos(θ)x+1))dx =−2π−cos(θ)[ln[x^2 −2cos(θ)x+1⌉_0 ^π +2sin^2 θ∫_0 ^π (1/((x−cos(θ))^2 +sin^2 (θ)))dx =2π−cos(θ)ln(π2−2cos(θ)π+1)+2∫_0 ^π (1/(((x/(sin(θ)))−cot (θ))^2 +1))dx if sin≠0 =−2π−cos(θ)ln(π^2 −2cos(θ)π+1)+2sin(θ){tan^(−1) {(x/(sinθ))−cot (θ)}_0 ^π } =−2π−cos(θ)ln{π^2 −2cos(θ)π+1}+2sin(θ){tan^(−1) {(π/(sin(θ)))−cot (θ)}−tan^(−1) {cot (−θ)} =−2π−cos(θ)ln{π^2 −2cos(θ)π+1}+2sin(θ){arctan{(π/(sin(θ)))−cot(θ)}−arctan(tg((π/2)+θ)) A(θ)=−2π+(π−cos(θ))ln(π^2 −2cos(θ)x+1)+2sin(θ)arctan(((π−cos(θ))/(sin(θ))))−2sin(θ)arctan(tan((π/2)+θ)) arctan(tg(a))=a if a∈]((−π)/2).(π/2)[ if a∈]−(π/2)+kπ.(π/2)+kπ[ arctantg(a)=a−kπ

$${let}\:{A}\left(\theta\right)=\int_{\mathrm{0}} ^{\pi} {ln}\left({x}^{\mathrm{2}} −\mathrm{2}{xcos}\left(\theta\right)+\mathrm{1}\right){d}\theta \\ $$$$\theta\in{IR}\backslash\left\{\mathrm{2}{k}\pi,{k}\in{IZ}\right\} \\ $$$${A}\left(\theta\right)=\left[{xln}\left({x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\pi} −\int\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{xcos}\left(\theta\right)}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}{dx} \\ $$$$=\pi{ln}\left(\pi^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right)\pi+\mathrm{1}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} −{xcos}\left(\theta\right)}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}{dx} \\ $$$$\frac{{x}^{\mathrm{2}} −{xcos}\left(\theta\right)}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}=\mathrm{1}+\frac{{xcos}\left(\theta\right)−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}} \\ $$$${A}\left(\theta\right)=\pi{ln}\left(\pi^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right)\pi+\mathrm{1}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \mathrm{1}+\frac{{xcos}\left(\theta\right)−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}{dx} \\ $$$${the}\:\mathrm{2}\:{nd}\:{integral}\: \\ $$$$=−\mathrm{2}\pi\:−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \frac{{xcos}\left(\theta\right)−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}{dx}=−\mathrm{2}\pi−\int_{\mathrm{0}} ^{\pi} \frac{{cos}\left(\theta\right)\left[\mathrm{2}{x}−\mathrm{2}{cos}\left(\theta\right)\right]−\mathrm{2}+\mathrm{2}{cos}^{\mathrm{2}} \left(\theta\right)}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}{dx} \\ $$$$=−\mathrm{2}\pi−{cos}\left(\theta\right)\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}{x}−\mathrm{2}{cos}\left(\theta\right)}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}{dx}+\mathrm{2}\int\frac{{sin}^{\mathrm{2}} \left(\theta\right)}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}}{dx} \\ $$$$=−\mathrm{2}\pi−{cos}\left(\theta\right)\left[{ln}\left[{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}\rceil_{\mathrm{0}} ^{\pi} +\mathrm{2}{sin}^{\mathrm{2}} \theta\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\left({x}−{cos}\left(\theta\right)\right)^{\mathrm{2}} +\mathrm{sin}\:^{\mathrm{2}} \left(\theta\right)}{dx}\right.\right. \\ $$$$=\mathrm{2}\pi−{cos}\left(\theta\right){ln}\left(\pi\mathrm{2}−\mathrm{2}{cos}\left(\theta\right)\pi+\mathrm{1}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\left(\frac{{x}}{\mathrm{sin}\left(\theta\right)}−\mathrm{cot}\:\left(\theta\right)\right)^{\mathrm{2}} +\mathrm{1}}{dx}\:{if}\:{sin}\neq\mathrm{0} \\ $$$$=−\mathrm{2}\pi−{cos}\left(\theta\right){ln}\left(\pi^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right)\pi+\mathrm{1}\right)+\mathrm{2}{sin}\left(\theta\right)\left\{\mathrm{tan}^{−\mathrm{1}} \left\{\frac{{x}}{{sin}\theta}−\mathrm{cot}\:\left(\theta\right)\right\}_{\mathrm{0}} ^{\pi} \right\} \\ $$$$=−\mathrm{2}\pi−{cos}\left(\theta\right){ln}\left\{\pi^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right)\pi+\mathrm{1}\right\}+\mathrm{2}{sin}\left(\theta\right)\left\{\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\pi}{{sin}\left(\theta\right)}−\mathrm{cot}\:\left(\theta\right)\right\}−\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(−\theta\right)\right\}\right. \\ $$$$=−\mathrm{2}\pi−{cos}\left(\theta\right){ln}\left\{\pi^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right)\pi+\mathrm{1}\right\}+\mathrm{2}{sin}\left(\theta\right)\left\{{arctan}\left\{\frac{\pi}{{sin}\left(\theta\right)}−{cot}\left(\theta\right)\right\}−{arctan}\left({tg}\left(\frac{\pi}{\mathrm{2}}+\theta\right)\right)\right. \\ $$$${A}\left(\theta\right)=−\mathrm{2}\pi+\left(\pi−{cos}\left(\theta\right)\right){ln}\left(\pi^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){x}+\mathrm{1}\right)+\mathrm{2}{sin}\left(\theta\right){arctan}\left(\frac{\pi−{cos}\left(\theta\right)}{{sin}\left(\theta\right)}\right)−\mathrm{2}{sin}\left(\theta\right){arctan}\left({tan}\left(\frac{\pi}{\mathrm{2}}+\theta\right)\right) \\ $$$$\left.{arctan}\left({tg}\left({a}\right)\right)={a}\:{if}\:{a}\in\right]\frac{−\pi}{\mathrm{2}}.\frac{\pi}{\mathrm{2}}\left[\right. \\ $$$$\left.{if}\:{a}\in\right]−\frac{\pi}{\mathrm{2}}+{k}\pi.\frac{\pi}{\mathrm{2}}+{k}\pi\left[\right. \\ $$$${arctantg}\left({a}\right)={a}−{k}\pi \\ $$$$ \\ $$

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