Question Number 32343 by abdo imad last updated on 23/Mar/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{{x}−{e}^{{it}} }\:\:. \\ $$
Commented by abdo imad last updated on 28/Mar/18
![let put I(x) = ∫_0 ^(2π) (dt/(x−e^(it) )) let suppose x≠0 I(x) = (1/x) ∫_0 ^(2π) (dt/(1−x^(−1) e^(it) )) case 1 ∣x^(−1) e^(it) ∣ <1 ⇔ ∣x∣>1⇒ x I(x) = ∫_0 ^(2π) ( Σ_(n=0) ^∞ x^(−n) e^(int) )dt =Σ_(n=0) ^∞ x^(−n) ∫_0 ^(2π) e^(int) dt =2π +Σ_(n=1) ^∞ x^(−n) [(1/(in)) e^(int) ]_0 ^(2π) =2π +0 =2π ⇒ I(x) =((2π)/x) . case 2 ∣x^(−1) e^(it) ∣>1 ⇔ ∣x∣<1 the ch.x=(1/α) ⇒∣α∣>1 I(x) =I((1/α)) = ∫_0 ^(2π) (dt/((1/α) −e^(it) )) = α ∫_0 ^(2π) (dt/(1−α e^(it) )) I(x) =∫_0 ^(2π) (dt/(x −cost −isint))dt = ∫_0 ^(2π) ((x−cost +isint)/((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 =_(t=π +u) ∫_(−π) ^π ((x +cosu)/((x+cosu)^2 +sin^2 u))du −i ∫_(−π) ^π ((sinu)/((x +cosu)^2 +sin^2 u))du = 2 ∫_0 ^π ((x +cosu)/(x^2 +2x cosu +1))du +0 .ch tan((u/2)) =t give I(x) = 2 ∫_0 ^(+∞) ((x +((1−t^2 )/(1+t^2 )))/(x^2 +2x ((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = 4∫_0 ^(+∞) ((x(1+t^2 ) +1−t^2 )/(x^2 (1+t^2 ) +2x(1−t^2 ))) dt =4 ∫_0 ^(+∞) ((x+1 +(x−1)t^2 )/((x^2 +2x +(x^2 −2x)t^2 )(1+t^2 )))dt =(4/x) ∫_0 ^(+∞) (((x−1)t^2 +x+1)/((x+2 +(x−2)t^2 )(1+t^2 )))dt ....be continued....](https://www.tinkutara.com/question/Q32568.png)
$${let}\:{put}\:{I}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{x}−{e}^{{it}} }\:{let}\:{suppose}\:{x}\neq\mathrm{0} \\ $$$${I}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{\mathrm{1}−{x}^{−\mathrm{1}} \:{e}^{{it}} } \\ $$$${case}\:\mathrm{1}\:\:\mid{x}^{−\mathrm{1}} {e}^{{it}} \mid\:<\mathrm{1}\:\:\Leftrightarrow\:\mid{x}\mid>\mathrm{1}\Rightarrow \\ $$$${x}\:{I}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{−{n}} \:{e}^{{int}} \right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{−{n}} \:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{e}^{{int}} {dt}\:\:=\mathrm{2}\pi\:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{−{n}} \left[\frac{\mathrm{1}}{{in}}\:{e}^{{int}} \right]_{\mathrm{0}} ^{\mathrm{2}\pi} \:=\mathrm{2}\pi\:+\mathrm{0} \\ $$$$=\mathrm{2}\pi\:\Rightarrow\:{I}\left({x}\right)\:=\frac{\mathrm{2}\pi}{{x}}\:. \\ $$$${case}\:\mathrm{2}\:\mid{x}^{−\mathrm{1}} {e}^{{it}} \mid>\mathrm{1}\:\Leftrightarrow\:\mid{x}\mid<\mathrm{1}\:\:{the}\:{ch}.{x}=\frac{\mathrm{1}}{\alpha}\:\Rightarrow\mid\alpha\mid>\mathrm{1} \\ $$$${I}\left({x}\right)\:={I}\left(\frac{\mathrm{1}}{\alpha}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{\frac{\mathrm{1}}{\alpha}\:−{e}^{{it}} }\:=\:\alpha\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{\mathrm{1}−\alpha\:{e}^{{it}} } \\ $$$${I}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{{x}\:−{cost}\:−{isint}}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{x}−{cost}\:+{isint}}{\left({x}−{cost}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {t}}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{x}−{cost}}{\left({x}−{cost}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {t}}{dt}\:\:+{i}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\left({x}−{cost}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {t}}{dt} \\ $$$$=_{{t}=\pi\:+{u}} \:\:\:\int_{−\pi} ^{\pi} \:\:\:\frac{{x}\:+{cosu}}{\left({x}+{cosu}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {u}}{du}\:−{i}\:\int_{−\pi} ^{\pi} \:\:\:\:\frac{{sinu}}{\left({x}\:+{cosu}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {u}}{du} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}\:+{cosu}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:{cosu}\:+\mathrm{1}}{du}\:+\mathrm{0}\:.{ch}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)\:={t}\:{give} \\ $$$${I}\left({x}\right)\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{x}\:+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\mathrm{4}\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+\mathrm{1}−{t}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+\mathrm{2}{x}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\:{dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\frac{{x}+\mathrm{1}\:+\left({x}−\mathrm{1}\right){t}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\left({x}^{\mathrm{2}} \:−\mathrm{2}{x}\right){t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\frac{\mathrm{4}}{{x}}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\left({x}−\mathrm{1}\right){t}^{\mathrm{2}} \:+{x}+\mathrm{1}}{\left({x}+\mathrm{2}\:+\left({x}−\mathrm{2}\right){t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:….{be}\:{continued}…. \\ $$
Answered by sma3l2996 last updated on 25/Mar/18
$${let}\:{z}={e}^{{it}} \Rightarrow{dt}=\frac{{dz}}{{iz}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{{x}−{e}^{{it}} }=\frac{\mathrm{1}}{{i}}\int_{\mid{z}\mid\leqslant\mathrm{1}} \frac{{dz}}{{z}\left({x}−{z}\right)} \\ $$$${if}\:\:{x}>\mathrm{1}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{{x}−{e}^{{it}} }=\mathrm{2}\pi{Res}\left({f};\mathrm{0}\right)=\frac{\mathrm{2}\pi}{{x}} \\ $$$${if}\:\:{x}\leqslant\mathrm{1}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{{x}−{e}^{{it}} }=\mathrm{2}\pi\left({Res}\left({f};\mathrm{0}\right)+{Res}\left({f};{x}\right)\right)=\mathrm{2}\pi\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{0} \\ $$