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Question Number 32483 by prof Abdo imad last updated on 25/Mar/18
calculate ∫_0 ^(2π) (dx/(1+2cosx)) .
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}+\mathrm{2}{cosx}}\:. \\ $$
Commented by prof Abdo imad last updated on 26/Mar/18
let put I = ∫_0 ^(2π) (dx/(1+2cosx)) the ch. x=π +t give I = ∫_(−π) ^π (dt/(1−2cost)) = 2 ∫_0 ^π (dt/(1−2cost)) and the ch. tan((t/2))=u hive I = 2 ∫_0 ^(+∞) (1/(1−2((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = 4 ∫_0 ^∞ (du/(1+u^2 −2 +2u^2 )) = 4∫_0 ^∞ (du/(3u^2 −1)) = 4 ∫_0 ^∞ (du/(((√3) u +1)((√3) −1))) =2 ∫_0 ^∞ ( (1/( (√3) t −1)) −(1/( (√3) t+1)))du =(2/( (√3)))[ ln∣(((√3)t−1)/( (√3)t +1))∣]_0 ^(+∞) =0 I =0
$${let}\:{put}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\mathrm{1}+\mathrm{2}{cosx}}\:{the}\:{ch}.\:{x}=\pi\:+{t}\:\:{give} \\ $$$${I}\:\:=\:\int_{−\pi} ^{\pi} \:\:\:\:\:\:\frac{{dt}}{\mathrm{1}−\mathrm{2}{cost}}\:\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{\mathrm{1}−\mathrm{2}{cost}}\:\:{and}\:{the} \\ $$$${ch}.\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{hive} \\ $$$${I}\:\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:−\mathrm{2}\:+\mathrm{2}{u}^{\mathrm{2}} }\:=\:\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\mathrm{3}{u}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left(\sqrt{\mathrm{3}}\:{u}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:{t}\:−\mathrm{1}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:{t}+\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left[\:{ln}\mid\frac{\sqrt{\mathrm{3}}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}{t}\:+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{0} \\ $$$${I}\:=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 26/Mar/18
method of residus ch. e^(ix) =z give I =∫_(∣z∣=1) (1/(1+2 ((z+z^(−1) )/2))) (dz/(iz)) = ∫_(∣z∣=1) (dz/(iz(1+z+z^(−1) ))) =∫_(∣z∣=1) ((−idz)/(z +z^2 +1)) = ∫_(∣z∣=1) ((−idz)/(z^2 +z +1)) let imtroduce the complex function ϕ(z)= ((−idz)/(z^2 +z +1)) .poles of ϕ? z^2 +z +1=0 ⇒ Δ=1−4=−3=(i(√3))^2 z_1 =((−1 +i(√3))/2) =j and z_2 =((−1−i(√3))/2) =j^− ∣z_1 ∣=1 and ∣z_2 ∣=1 so ∫_(∣z∣=1) ϕ(z)dz =2iπ( Res(ϕ,z_1 ) +Res(ϕ,z_2 )) but we have ϕ(z) = ((−i)/((z−z_1 )(z−z_2 ))) Res(ϕ,z_1 ) = ((−i)/(z_1 −z_2 )) = ((−i)/(i(√3))) = ((−1)/( (√3))) Res(ϕ,z_2 ) = ((−i)/(z_2 −z_1 )) = ((−i)/(−i(√3))) = (1/( (√3))) ∫_(∣z∣=1) ϕ(z)dz = 2iπ( ((−1)/( (√3))) +(1/( (√3))))=0 ⇒ I =0
$${method}\:{of}\:{residus}\:{ch}.\:{e}^{{ix}} ={z}\:{give} \\ $$$${I}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{{dz}}{{iz}\left(\mathrm{1}+{z}+{z}^{−\mathrm{1}} \right)}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−{idz}}{{z}\:+{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−{idz}}{{z}^{\mathrm{2}} \:+{z}\:+\mathrm{1}}\:{let}\:{imtroduce}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)=\:\:\:\frac{−{idz}}{{z}^{\mathrm{2}} \:+{z}\:+\mathrm{1}}\:.{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} \:+{z}\:+\mathrm{1}=\mathrm{0}\:\Rightarrow\:\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\frac{−\mathrm{1}\:+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={j}\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\overset{−} {{j}} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\mathrm{1}\:{and}\:\mid{z}_{\mathrm{2}} \mid=\mathrm{1}\:{so} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:+{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\right) \\ $$$${but}\:{we}\:{have}\:\varphi\left({z}\right)\:=\:\frac{−{i}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:\frac{−{i}}{{z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} }\:=\:\frac{−{i}}{{i}\sqrt{\mathrm{3}}}\:=\:\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\:\frac{−{i}}{{z}_{\mathrm{2}} \:−{z}_{\mathrm{1}} }\:=\:\frac{−{i}}{−{i}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\left(\:\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{0}\:\:\Rightarrow\:{I}\:=\mathrm{0} \\ $$

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