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Question Number 39633 by abdo mathsup 649 cc last updated on 09/Jul/18
find the value of f(x) = ∫_0 ^π ln(x^2 −2x cosθ +1)dθ with x fromR.
$${find}\:{the}\:{value}\:{of}\: \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\pi} {ln}\left({x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cos}\theta\:+\mathrm{1}\right){d}\theta\:\:{with}\:{x}\:{fromR}. \\ $$
Commented by math khazana by abdo last updated on 09/Jul/18
we have f^′ (x)= ∫_0 ^π ((2x−2cosθ)/(x^(2 ) −2xcosθ +1))dθ =_(θ=2t) ∫_0 ^(2π) ((2x−2cos(2t))/(x^2 −2xcos(2t) +1)) 2dt = 4 ∫_0 ^(2π) ((x−cos(2t))/(x^2 −2xcos(2t)+1))dt changement e^(it) =z give f^′ (x)=4 ∫_(∣z∣=1) ((x−((z^2 +z^(−2) )/2))/(x^2 −2x ((z^2 +z^(−2) )/2) +1)) (dz/(iz)) =4 ∫_(∣z∣=1) ((2x−z^2 −z^(−2) )/(iz(x^2 −xz^2 −xz^(−2) +1)))dz =4 ∫_(∣z∣=1) ((2xz^2 −z^4 −1)/(iz(x^2 z^2 −x z^4 −x +z^2 )))dz =−4i ∫_(∣z∣=1) ((−z^4 +2xz^2 −1)/(z(−xz^4 +(1+x^2 )z^2 −x)))dz =−4i ∫_(∣z∣=1) ((z^4 −2xz^2 +1)/(z{xz^4 −(1+x^2 )z^2 +x}))dz let ϕ(z) = ((z^4 −2xz^2 +1)/(z{ xz^4 −(1+x^2 )z^2 +x})) .poles of ϕ roots of xz^4 −(1+x^2 )z^2 +x Δ =(1+x^2 )^2 −4x^2 =1+2x^2 +x^4 −4x^2 =1−2x^2 +x^4 =(1−x^2 )^2 ⇒ z^2 =((1+x^2 +∣1−x^2 ∣)/(2x)) and z^2 =((1+x^2 −∣1−x^2 ∣)/(2x)) (we suppose that x≠0) case 1 ∣x∣<1 ⇒z^2 =((1+x^2 +1−x^2 )/(2x)) = (1/x) or z^2 =((1+x^2 −1+x^2 )/(2x)) =x so if 0<x<1 we get z =+^− (1/( (√x))) and z =+^− (√x) ,0 are poles of ϕ ϕ(z) = ((z^4 −2xz^2 +1)/(xz( z−(1/( (√x))))(z+(1/( (√x))))(z−(√x))(z+(√x)))) ∫_(∣z∣=1) ϕ(z)dz =2iπ { Res(ϕ,0) +Res(ϕ,(√x)) +Res(ϕ,−(√x))} Res(ϕ,0) = (1/x) Res(ϕ,(√x)) = ((x^2 −2x^2 +1)/(x(√x)(x−(1/x))2(√x))) =((1−x^2 )/(2x^2 (((x^2 −1)/x)))) = ((−1)/(2x)) Res(ϕ,−(√x)) =((x^2 −2x^2 +1)/(−x(√x)(x−(1/x))(−2(√x)))) =((1−x^2 )/(2x^2 (((x^2 −1)/x)))) =((−1)/(2x)) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ{ (1/x) −(1/(2x)) −(1/(2x))}=0 ⇒ f^′ (x) =0 ⇒ f(x)=c = f(0)=0 and we get the same result if −1<x<0 case 2 ∣x∣>1 we have f(x) = ∫_0 ^π ln(x^2 (1 −(2/x) cosθ + (1/x^2 )))dθ =2π ln∣x∣ + ∫_0 ^π ln( X^2 −2X cosθ +1)dθ =2π ln∣x∣ +0 because ∣X∣ = (1/(∣x∣))<1 ⇒ f(x) =2πln∣x∣ if ∣x∣>1 and f(x)=0 if∣x∣<1 .
$${we}\:{have}\:\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{2}{x}−\mathrm{2}{cos}\theta}{{x}^{\mathrm{2}\:} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}}{d}\theta \\ $$$$=_{\theta=\mathrm{2}{t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{\mathrm{2}{x}−\mathrm{2}{cos}\left(\mathrm{2}{t}\right)}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\left(\mathrm{2}{t}\right)\:+\mathrm{1}}\:\mathrm{2}{dt} \\ $$$$=\:\mathrm{4}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{x}−{cos}\left(\mathrm{2}{t}\right)}{{x}^{\mathrm{2}} \:\:−\mathrm{2}{xcos}\left(\mathrm{2}{t}\right)+\mathrm{1}}{dt}\:\:{changement} \\ $$$${e}^{{it}} ={z}\:{give} \\ $$$${f}^{'} \left({x}\right)=\mathrm{4}\:\int_{\mid{z}\mid=\mathrm{1}} \:\frac{{x}−\frac{{z}^{\mathrm{2}} \:+{z}^{−\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:\frac{{z}^{\mathrm{2}} \:+{z}^{−\mathrm{2}} }{\mathrm{2}}\:+\mathrm{1}}\:\frac{{dz}}{{iz}} \\ $$$$=\mathrm{4}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{2}{x}−{z}^{\mathrm{2}} \:−{z}^{−\mathrm{2}} }{{iz}\left({x}^{\mathrm{2}} \:−{xz}^{\mathrm{2}} \:−{xz}^{−\mathrm{2}} +\mathrm{1}\right)}{dz} \\ $$$$=\mathrm{4}\:\int_{\mid{z}\mid=\mathrm{1}} \:\frac{\mathrm{2}{xz}^{\mathrm{2}} \:−{z}^{\mathrm{4}} \:−\mathrm{1}}{{iz}\left({x}^{\mathrm{2}} \:{z}^{\mathrm{2}} \:−{x}\:{z}^{\mathrm{4}} \:−{x}\:+{z}^{\mathrm{2}} \right)}{dz} \\ $$$$=−\mathrm{4}{i}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−{z}^{\mathrm{4}} \:+\mathrm{2}{xz}^{\mathrm{2}} \:−\mathrm{1}}{{z}\left(−{xz}^{\mathrm{4}} \:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:−{x}\right)}{dz} \\ $$$$=−\mathrm{4}{i}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{z}^{\mathrm{4}} \:−\mathrm{2}{xz}^{\mathrm{2}} \:+\mathrm{1}}{{z}\left\{{xz}^{\mathrm{4}} \:−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:+{x}\right\}}{dz} \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{4}} \:−\mathrm{2}{xz}^{\mathrm{2}} \:+\mathrm{1}}{{z}\left\{\:{xz}^{\mathrm{4}} \:−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:+{x}\right\}}\:.{poles}\:{of}\:\varphi \\ $$$${roots}\:{of}\:\:{xz}^{\mathrm{4}} \:−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:+{x}\: \\ $$$$\Delta\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{4}{x}^{\mathrm{2}} \:=\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:−\mathrm{4}{x}^{\mathrm{2}} \\ $$$$=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:\Rightarrow \\ $$$${z}^{\mathrm{2}} \:=\frac{\mathrm{1}+{x}^{\mathrm{2}} \:+\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}{\mathrm{2}{x}}\:\:\:{and}\:\:{z}^{\mathrm{2}} \:=\frac{\mathrm{1}+{x}^{\mathrm{2}} \:−\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}{\mathrm{2}{x}} \\ $$$$\left({we}\:{suppose}\:{that}\:{x}\neq\mathrm{0}\right) \\ $$$${case}\:\mathrm{1}\:\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{z}^{\mathrm{2}} \:\:=\frac{\mathrm{1}+{x}^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}}\:=\:\frac{\mathrm{1}}{{x}} \\ $$$${or}\:{z}^{\mathrm{2}} \:=\frac{\mathrm{1}+{x}^{\mathrm{2}} \:−\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}}\:={x}\:\:{so}\:{if}\:\:\:\mathrm{0}<{x}<\mathrm{1}\:{we}\:{get} \\ $$$${z}\:=\overset{−} {+}\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\:\:{and}\:\:{z}\:=\overset{−} {+}\:\sqrt{{x}}\:\:\:,\mathrm{0}\:{are}\:{poles}\:{of}\:\varphi \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{{z}^{\mathrm{4}} \:−\mathrm{2}{xz}^{\mathrm{2}} \:+\mathrm{1}}{{xz}\left(\:{z}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)\left({z}+\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)\left({z}−\sqrt{{x}}\right)\left({z}+\sqrt{{x}}\right)} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,\mathrm{0}\right)\:+{Res}\left(\varphi,\sqrt{{x}}\right)\right. \\ $$$$\left.+{Res}\left(\varphi,−\sqrt{{x}}\right)\right\} \\ $$$${Res}\left(\varphi,\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{{x}} \\ $$$${Res}\left(\varphi,\sqrt{{x}}\right)\:=\:\frac{{x}^{\mathrm{2}} \:−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}\sqrt{{x}}\left({x}−\frac{\mathrm{1}}{{x}}\right)\mathrm{2}\sqrt{{x}}}\:=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} \:−\mathrm{1}}{{x}}\right)} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{2}{x}} \\ $$$${Res}\left(\varphi,−\sqrt{{x}}\right)\:=\frac{{x}^{\mathrm{2}} \:−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{−{x}\sqrt{{x}}\left({x}−\frac{\mathrm{1}}{{x}}\right)\left(−\mathrm{2}\sqrt{{x}}\right)} \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} \:−\mathrm{1}}{{x}}\right)}\:=\frac{−\mathrm{1}}{\mathrm{2}{x}}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}}\right\}=\mathrm{0}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)={c}\:=\:{f}\left(\mathrm{0}\right)=\mathrm{0}\:{and}\:{we}\:{get}\:{the} \\ $$$${same}\:{result}\:{if}\:\:−\mathrm{1}<{x}<\mathrm{0} \\ $$$${case}\:\mathrm{2}\:\:\mid{x}\mid>\mathrm{1}\:{we}\:{have} \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left({x}^{\mathrm{2}} \left(\mathrm{1}\:−\frac{\mathrm{2}}{{x}}\:{cos}\theta\:\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right){d}\theta \\ $$$$=\mathrm{2}\pi\:{ln}\mid{x}\mid\:\:+\:\int_{\mathrm{0}} ^{\pi} \:\:{ln}\left(\:{X}^{\mathrm{2}} \:−\mathrm{2}{X}\:{cos}\theta\:+\mathrm{1}\right){d}\theta\: \\ $$$$=\mathrm{2}\pi\:{ln}\mid{x}\mid\:+\mathrm{0}\:\:{because}\:\mid{X}\mid\:=\:\frac{\mathrm{1}}{\mid{x}\mid}<\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\mathrm{2}\pi{ln}\mid{x}\mid\:{if}\:\:\mid{x}\mid>\mathrm{1}\:\:{and}\:{f}\left({x}\right)=\mathrm{0}\:{if}\mid{x}\mid<\mathrm{1}\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jul/18
x^2 −2xcosθ+1 (x−cosθ)^2 +sin^2 θ (x−cosθ+isinθ)(x−cosθ−isinθ) (x−e^(−iθ) )(x−e^(iθ) ) f(x)=∫_0 ^Π {ln(x−e^(−iθ) )+ln(x−e^(iθ) )}dθ (dI/dx)=∫_0 ^Π (1/(x−e^(−iθ) ))+(1/(x−e^(iθ) )) dθ =∫_0 ^Π (e^(iθ) /(xe^(iθ) −1))dθ+∫_0 ^Π (e^(−iθ) /(xe^(−iθ) −1))dθ =(1/x)∫_0 ^Π (e^(iθ) /(e^(iθ) −(1/x)))dθ+(1/x)∫_0 ^Π (e^(−iθ) /(e^(−iθ) −(1/x)))dθ =∣(1/(ix))ln(e^(iθ) −(1/x))+(1/(−ix))ln(e^(−iθ) −(1/x))∣_0 ^Π =(1/(ix))∣ln(((e^(iθ) −(1/x))/(e^(−iθ) −(1/x))))∣_0 ^Π e^(iΠ) =cosΠ+isinΠ=−1 (1/(ix)){ln(((−1−(1/x))/(−1−(1/x))))−ln(((1−(1/x))/(1−(1/x))))}=0 (dI/dx)=0 I=constant pls check my steps...whether any mistake...
$${x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta+\mathrm{1} \\ $$$$\left({x}−{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta \\ $$$$\left({x}−{cos}\theta+{isin}\theta\right)\left({x}−{cos}\theta−{isin}\theta\right) \\ $$$$\left({x}−{e}^{−{i}\theta} \right)\left({x}−{e}^{{i}\theta} \right) \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\Pi} \left\{{ln}\left({x}−{e}^{−{i}\theta} \right)+{ln}\left({x}−{e}^{{i}\theta} \right)\right\}{d}\theta \\ $$$$\frac{{dI}}{{dx}}=\int_{\mathrm{0}} ^{\Pi} \frac{\mathrm{1}}{{x}−{e}^{−{i}\theta} }+\frac{\mathrm{1}}{{x}−{e}^{{i}\theta} }\:\:{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\Pi} \frac{{e}^{{i}\theta} }{{xe}^{{i}\theta} −\mathrm{1}}{d}\theta+\int_{\mathrm{0}} ^{\Pi} \frac{{e}^{−{i}\theta} }{{xe}^{−{i}\theta} −\mathrm{1}}{d}\theta \\ $$$$=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\Pi} \frac{{e}^{{i}\theta} }{{e}^{{i}\theta} −\frac{\mathrm{1}}{{x}}}{d}\theta+\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\Pi} \frac{{e}^{−{i}\theta} }{{e}^{−{i}\theta} −\frac{\mathrm{1}}{{x}}}{d}\theta \\ $$$$=\mid\frac{\mathrm{1}}{{ix}}{ln}\left({e}^{{i}\theta} −\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{1}}{−{ix}}{ln}\left({e}^{−{i}\theta} −\frac{\mathrm{1}}{{x}}\right)\mid_{\mathrm{0}} ^{\Pi} \\ $$$$=\frac{\mathrm{1}}{{ix}}\mid{ln}\left(\frac{{e}^{{i}\theta} −\frac{\mathrm{1}}{{x}}}{{e}^{−{i}\theta} −\frac{\mathrm{1}}{{x}}}\right)\mid_{\mathrm{0}} ^{\Pi} \:\:\:{e}^{{i}\Pi} ={cos}\Pi+{isin}\Pi=−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{ix}}\left\{{ln}\left(\frac{−\mathrm{1}−\frac{\mathrm{1}}{{x}}}{−\mathrm{1}−\frac{\mathrm{1}}{{x}}}\right)−{ln}\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\right)\right\}=\mathrm{0} \\ $$$$\frac{{dI}}{{dx}}=\mathrm{0} \\ $$$${I}={constant} \\ $$$${pls}\:{check}\:{my}\:{steps}…{whether}\:{any}\:{mistake}… \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 09/Jul/18
sir Tanmay the function ln at C is not like ln at R look ln(−1)in R don t exist but in C ln(−1)=ln(e^(iπ) )=iπ so you have commited a error in the final lines you must extract Re (∫) and Im( ∫) to have a correct answer ...
$${sir}\:{Tanmay}\:\:{the}\:{function}\:{ln}\:\:{at}\:\:{C}\:{is}\:{not}\:{like}\:{ln}\:{at}\:\:{R}\:\:{look}\:\:{ln}\left(−\mathrm{1}\right){in}\:{R} \\ $$$${don}\:{t}\:{exist}\:{but}\:\:{in}\:\:{C}\:{ln}\left(−\mathrm{1}\right)={ln}\left({e}^{{i}\pi} \right)={i}\pi\:\:{so}\:{you}\:{have}\:{commited}\:{a}\:{error} \\ $$$${in}\:{the}\:{final}\:{lines}\:\:{you}\:{must}\:{extract}\:\:{Re}\:\left(\int\right)\:{and}\:{Im}\left(\:\int\right)\:{to}\:{have}\:{a}\:{correct} \\ $$$${answer}\:… \\ $$

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