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Question Number 36931 by maxmathsup by imad last updated on 07/Jun/18
calculate ∫_0 ^(2π) (dt/(x −e^(it) ))
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{x}\:−{e}^{{it}} } \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
let f(x) = ∫_0 ^(2π) (dt/(x −e^(it) )) f(x)= ∫_0 ^(2π) (dt/(x−cost −isint)) =∫_0 ^(2π) ((x−cost +i sint)/((x−cost)^2 +sin^2 t))dt =∫_0 ^(2π) ((x−cost)/((x−cost)^2 +sin^2 t))dt +i ∫_0 ^(2π) ((sint)/((x−cost)^2 +sin^2 t))dt =I +iJ I = ∫_0 ^π ((x−cost)/(x^2 −2x cost +1))dt +∫_π ^(2π) ((x−cost)/(x^2 −2xcost +1)) chang.tan((t/2)) =u give ∫_0 ^π ((x−cost)/(x^2 −2x cost +1))dt = ∫_0 ^∞ ((x −((1−u^2 )/(1+u^2 )))/(x^2 −2x ((1−u^2 )/(1+u^2 )) +1)) ((2du)/(1+u^2 )) = ∫_0 ^∞ ((x(1+u^2 ) −1+u^2 )/((1+u^2 )^2 (((x^2 (1+u^2 )−2x+2xu^2 )/(1+u^2 )))))du =∫_0 ^∞ ((x +xu^2 −1+u^2 )/((1+u^2 ){x^2 +x^2 u^2 −2x +2xu^2 )))du =∫_0 ^∞ (((x+1)u^2 +x−1)/((1+u^2 ){ (x^2 +2x)u^2 +x^2 −2x}))du =(1/x)∫_0 ^∞ (((x+1)u^2 +x−1)/((1+u^2 )( (x+2)u^2 +x−2))) du = (1/(x(x+2)))∫_0 ^∞ (((x+1)u^2 +x−1)/((1+u^2 )(u^2 +((x−2)/(x+2)))))du case1 ((x−2)/(x+2))>0 andx(x+2)≠0 let ϕ(z)= (((x+1)z^2 +x−1)/((1+z^2 )(z^(2 ) +((x−2)/(x+2))))) the poles of ϕ are i ,−i, i(√((x−2)/(x+2))),−i(√((x−2)/(x+2))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i)+Res(ϕ,i(√((x−2)/(x+2))))} Res(ϕ,i) = (((x+1)(−1) +x−1)/(2i(−1 +((x−2)/(x+2))))) =((−2)/(2i((−4)/(x+2)))) =((−2(x+2))/(−8i)) = ((x+2)/(4i)) Res(ϕ,i(√((x−2)/(x+2)))) = ((−(x+1)((x−2)/(x+2)) +x−1)/(2i(√((x−2)/(x+2)))(1−((x−2)/(x+2))))) = (((x−1)(x+2)−(x+1)(x−2))/(2i(x+2)(√((x−2)/(x+2)))(x+2 −x+2))) = ((x^2 +x −2 −(x^2 −x−2))/(8i(x+2)(√((x−2)/(x+2))))) = ((2x)/(8i(√(x^2 −4)))) = (x/(4i(√(x^2 −4)))) ∫_(−∞) ^+ ϕ(z)dz =2iπ{ ((x+2)/(4i)) + (x/(4i(√(x^2 −4))))} =(π/2)(x+2) +((πx)/(2(√(x^2 −4)))) ⇒ ∫_0 ^π ((x−cost)/(x^2 −2x cost +1)) dt =(1/(x(x+2))){(π/4)(x+2) +((πx)/(4(√(x^2 −4)))) .} = (π/(4x)) +(π/(4(x+2)(√(x^2 −4)))) . case 2 ((x−2)/(x+2))<0 ⇒ ϕ(z) = (((x+1)z^2 +x−1)/((1+z^2 )(z^2 − ((2−x)/(2+x))))) and the roots of ϕ are i,−i,(√((2−x)/(2+x))) ,−(√( ((2−x)/(2+x)))) and we follow the same method...
$${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{x}\:−{e}^{{it}} } \\ $$$${f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{x}−{cost}\:−{isint}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{x}−{cost}\:+{i}\:{sint}}{\left({x}−{cost}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {t}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{x}−{cost}}{\left({x}−{cost}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {t}}{dt}\:+{i}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{sint}}{\left({x}−{cost}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {t}}{dt} \\ $$$$={I}\:+{iJ} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}−{cost}}{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cost}\:\:+\mathrm{1}}{dt}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\:\:\frac{{x}−{cost}}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcost}\:+\mathrm{1}} \\ $$$${chang}.{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}−{cost}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:{cost}\:+\mathrm{1}}{dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}\:−\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:+\mathrm{1}}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\:−\mathrm{1}+{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \:\left(\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)−\mathrm{2}{x}+\mathrm{2}{xu}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{x}\:+{xu}^{\mathrm{2}} \:−\mathrm{1}+{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left\{{x}^{\mathrm{2}} \:+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \:−\mathrm{2}{x}\:+\mathrm{2}{xu}^{\mathrm{2}} \right)}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left({x}+\mathrm{1}\right){u}^{\mathrm{2}} \:+{x}−\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left\{\:\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\right){u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \:−\mathrm{2}{x}\right\}}{du} \\ $$$$=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left({x}+\mathrm{1}\right){u}^{\mathrm{2}} \:+{x}−\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\:\left({x}+\mathrm{2}\right){u}^{\mathrm{2}} \:+{x}−\mathrm{2}\right)}\:{du} \\ $$$$=\:\frac{\mathrm{1}}{{x}\left({x}+\mathrm{2}\right)}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left({x}+\mathrm{1}\right){u}^{\mathrm{2}} \:+{x}−\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({u}^{\mathrm{2}} \:+\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}\right)}{du} \\ $$$${case}\mathrm{1}\:\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}>\mathrm{0}\:{andx}\left({x}+\mathrm{2}\right)\neq\mathrm{0}\:{let} \\ $$$$\varphi\left({z}\right)=\:\frac{\left({x}+\mathrm{1}\right){z}^{\mathrm{2}} \:+{x}−\mathrm{1}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)\left({z}^{\mathrm{2}\:} \:+\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}\right)}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\:,−{i},\:{i}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}},−{i}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,{i}\right)+{Res}\left(\varphi,{i}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}}\right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\frac{\left({x}+\mathrm{1}\right)\left(−\mathrm{1}\right)\:+{x}−\mathrm{1}}{\mathrm{2}{i}\left(−\mathrm{1}\:+\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}\right)}\:=\frac{−\mathrm{2}}{\mathrm{2}{i}\frac{−\mathrm{4}}{{x}+\mathrm{2}}} \\ $$$$=\frac{−\mathrm{2}\left({x}+\mathrm{2}\right)}{−\mathrm{8}{i}}\:=\:\frac{{x}+\mathrm{2}}{\mathrm{4}{i}} \\ $$$${Res}\left(\varphi,{i}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}}\right)\:=\:\frac{−\left({x}+\mathrm{1}\right)\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}\:+{x}−\mathrm{1}}{\mathrm{2}{i}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}}\left(\mathrm{1}−\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}\right)} \\ $$$$=\:\frac{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)−\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)}{\mathrm{2}{i}\left({x}+\mathrm{2}\right)\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}}\left({x}+\mathrm{2}\:−{x}+\mathrm{2}\right)} \\ $$$$=\:\frac{{x}^{\mathrm{2}} \:+{x}\:−\mathrm{2}\:−\left({x}^{\mathrm{2}} \:−{x}−\mathrm{2}\right)}{\mathrm{8}{i}\left({x}+\mathrm{2}\right)\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}}} \\ $$$$=\:\frac{\mathrm{2}{x}}{\mathrm{8}{i}\sqrt{{x}^{\mathrm{2}} \:−\mathrm{4}}}\:=\:\frac{{x}}{\mathrm{4}{i}\sqrt{{x}^{\mathrm{2}} \:−\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{{x}+\mathrm{2}}{\mathrm{4}{i}}\:+\:\frac{{x}}{\mathrm{4}{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left({x}+\mathrm{2}\right)\:+\frac{\pi{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:−\mathrm{4}}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}−{cost}}{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cost}\:+\mathrm{1}}\:{dt}\:=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{2}\right)}\left\{\frac{\pi}{\mathrm{4}}\left({x}+\mathrm{2}\right)\:+\frac{\pi{x}}{\mathrm{4}\sqrt{{x}^{\mathrm{2}} \:−\mathrm{4}}}\:.\right\} \\ $$$$=\:\frac{\pi}{\mathrm{4}{x}}\:+\frac{\pi}{\mathrm{4}\left({x}+\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} \:−\mathrm{4}}}\:. \\ $$$${case}\:\mathrm{2}\:\:\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}<\mathrm{0}\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\:\frac{\left({x}+\mathrm{1}\right){z}^{\mathrm{2}} \:+{x}−\mathrm{1}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)\left({z}^{\mathrm{2}} \:−\:\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}\right)}\:{and}\:{the}\:{roots}\:{of}\:\varphi\:{are} \\ $$$${i},−{i},\sqrt{\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}}\:\:,−\sqrt{\:\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}}\:\:{and}\:{we}\:{follow}\:{the}\:{same} \\ $$$${method}… \\ $$$$ \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
∫_π ^(2π) ((x−cost)/(x^2 −2xcost +1))dt =_(t=π+u) ∫_0 ^π ((x+cosu)/(x^2 +2x cosu +1))du ...
$$\int_{\pi} ^{\mathrm{2}\pi} \:\:\:\frac{{x}−{cost}}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcost}\:+\mathrm{1}}{dt}\:=_{{t}=\pi+{u}} \:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}+{cosu}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:{cosu}\:+\mathrm{1}}{du} \\ $$$$… \\ $$

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