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Question Number 200130 by universe last updated on 14/Nov/23
solve by contour integrstion ∫_0 ^(2π) (dx/(1+acosx))
$$\:\:{solve}\:{by}\:{contour}\:{integrstion} \\ $$$$\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{1}+{a}\mathrm{cos}{x}}\: \\ $$
Answered by Mathspace last updated on 14/Nov/23
I=∫_0 ^(2π) (dx/(1+a((e^(ix) +e^(−ix) )/2))) (e^(ix) =z) =∫_(∣z∣=1) (dz/(iz(1+a((z+z^(−1) )/2)))) =∫_(∣z∣=1) ((−2idz)/(z(2+az+az^(−1) ))) =∫_(∣z∣=1) ((−2idz)/(2z+az^2 +a)) =∫_(∣z∣=1) ((−2idz)/(az^2 +2z+a)) f(z)=((−2i)/(az^2 +2z+a)) poles? Δ^′ =1−a^2 case 1 ∣a∣<1 ⇒z_1 =((−1+(√(1−a^2 )))/a) z_2 =((−1−(√(1−a^2 )))/a) (a≠0) ∣z_1 ∣−1=((1−(√(1−a^2 )))/(∣a∣))−1 =((1−(√(1−a^2 ))−∣a∣)/(∣a∣))=((1−∣a∣−(√(1−a^2 )))/(∣a∣)) (1−∣a∣)^2 −(1−a^2 ) =1−2∣a∣+a^2 −1+a^2 =2a^2 −2∣a∣=2∣a∣(∣a∣−1)<0 ⇒∣z_1 ∣<1 ∣z_2 ∣−1=((1+(√(1−a^2 )))/(∣a∣))−1 =((1−∣a∣+(√(1−a^2 )))/(∣a∣))>0 ⇒∣z_2 ∣>1 f(z)=((−2i)/(a(z−z_1 )(z−z_2 ))) ∫_(∣z∣=1) f(z)dz=2iπRes(f,z_1 ) =2iπ.((−2i)/(a(z_1 −z_2 )))=((4π)/(a(2((√(1−a^2 ))/a)))) =((2π)/( (√(1−a^2 )))) ⇒ I=((2π)/( (√(1−a^2 )))) rest the case ⇒∣a∣>1 z_1 =((−1+i(√(a^2 −1)))/a) and z_2 =((−1−i(√(a^2 −1)))/a) ....follow the same path...
$${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{1}+{a}\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}}\:\:\left({e}^{{ix}} ={z}\right) \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{{dz}}{{iz}\left(\mathrm{1}+{a}\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{{z}\left(\mathrm{2}+{az}+{az}^{−\mathrm{1}} \right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{\mathrm{2}{z}+{az}^{\mathrm{2}} +{a}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{{az}^{\mathrm{2}} +\mathrm{2}{z}+{a}} \\ $$$${f}\left({z}\right)=\frac{−\mathrm{2}{i}}{{az}^{\mathrm{2}} +\mathrm{2}{z}+{a}}\:\:{poles}? \\ $$$$\Delta^{'} =\mathrm{1}−{a}^{\mathrm{2}} \\ $$$${case}\:\mathrm{1}\:\:\mid{a}\mid<\mathrm{1}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{{a}} \\ $$$${z}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{{a}}\:\:\:\left({a}\neq\mathrm{0}\right) \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}=\frac{\mathrm{1}−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mid{a}\mid}−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }−\mid{a}\mid}{\mid{a}\mid}=\frac{\mathrm{1}−\mid{a}\mid−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mid{a}\mid} \\ $$$$\left(\mathrm{1}−\mid{a}\mid\right)^{\mathrm{2}} −\left(\mathrm{1}−{a}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}−\mathrm{2}\mid{a}\mid+{a}^{\mathrm{2}} −\mathrm{1}+{a}^{\mathrm{2}} \\ $$$$=\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}\mid{a}\mid=\mathrm{2}\mid{a}\mid\left(\mid{a}\mid−\mathrm{1}\right)<\mathrm{0} \\ $$$$\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mid{a}\mid}−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\mid{a}\mid+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mid{a}\mid}>\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{2}} \mid>\mathrm{1} \\ $$$${f}\left({z}\right)=\frac{−\mathrm{2}{i}}{{a}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f},{z}_{\mathrm{1}} \right) \\ $$$$=\mathrm{2}{i}\pi.\frac{−\mathrm{2}{i}}{{a}\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}=\frac{\mathrm{4}\pi}{{a}\left(\mathrm{2}\frac{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{{a}}\right)} \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${I}=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }} \\ $$$${rest}\:{the}\:{case}\:\Rightarrow\mid{a}\mid>\mathrm{1} \\ $$$${z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{{a}} \\ $$$${and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{{a}} \\ $$$$….{follow}\:{the}\:{same}\:{path}… \\ $$$$ \\ $$
Answered by MM42 last updated on 15/Nov/23
I=∫ (dx/(1+acosx)) if a=0⇒I=x+c if a=1⇒I=(1/2)∫ (1+tan^2 (x/2))dx⇒I= tan(x/2)+c if a=−1⇒I=(1/2)∫ (1+cotan^2 (x/2))dx⇒I= −cotan(x/2)+c otherwise ⇒let tan(x/2)=u⇒∫ ((2du)/((1+a)+(1−a)u^2 ))=(2/(1−a))∫ (du/(((1+a)/(1−a))+u^2 )) let ((1+a)/(1−a))=k ⇒I=(2/(1−a)) ∫ (du/(u^2 +k)) if ∣a∣<1 ⇒k>0⇒ I=(2/(1−a)) ×(1/( (√k)))×tan^(−1) ((u/( (√k))))+c =(2/( (√(1−a^2 ))))×tan^(−1) ((√((1−a)/(1+a))) tan((x/2)×(√((1−a)/(1+a)))))+c if ∣a∣>1 ⇒k<0⇒ I=(1/(1−a)) ×(1/( k))×ln(((u−k)/(u+k)))+c =(1/( (√(1−a^2 ))))×ln((((1−a)tan(x/2)−(√(1+a)))/((1−a)tan(x/2)+(√(1+a)))))+c
$${I}=\int\:\frac{{dx}}{\mathrm{1}+{acosx}} \\ $$$${if}\:{a}=\mathrm{0}\Rightarrow{I}={x}+{c} \\ $$$${if}\:{a}=\mathrm{1}\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right){dx}\Rightarrow{I}=\:{tan}\frac{{x}}{\mathrm{2}}+{c} \\ $$$${if}\:{a}=−\mathrm{1}\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\mathrm{1}+{cotan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right){dx}\Rightarrow{I}=\:−{cotan}\frac{{x}}{\mathrm{2}}+{c} \\ $$$${otherwise} \\ $$$$\Rightarrow{let}\:\:{tan}\frac{{x}}{\mathrm{2}}={u}\Rightarrow\int\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{a}\right)+\left(\mathrm{1}−{a}\right){u}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{1}−{a}}\int\:\frac{{du}}{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}+{u}^{\mathrm{2}} }\:\: \\ $$$${let}\:\:\:\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}={k}\:\Rightarrow{I}=\frac{\mathrm{2}}{\mathrm{1}−{a}}\:\int\:\:\frac{{du}}{{u}^{\mathrm{2}} +{k}} \\ $$$${if}\:\:\mid{a}\mid<\mathrm{1}\:\Rightarrow{k}>\mathrm{0}\Rightarrow\:{I}=\frac{\mathrm{2}}{\mathrm{1}−{a}}\:×\frac{\mathrm{1}}{\:\sqrt{{k}}}×{tan}^{−\mathrm{1}} \left(\frac{{u}}{\:\sqrt{{k}}}\right)+{c} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}×{tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}}}\:{tan}\left(\frac{{x}}{\mathrm{2}}×\sqrt{\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}}}\right)\right)+{c} \\ $$$$ \\ $$$${if}\:\:\mid{a}\mid>\mathrm{1}\:\Rightarrow{k}<\mathrm{0}\Rightarrow\:{I}=\frac{\mathrm{1}}{\mathrm{1}−{a}}\:×\frac{\mathrm{1}}{\:{k}}×{ln}\left(\frac{{u}−{k}}{{u}+{k}}\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}×{ln}\left(\frac{\left(\mathrm{1}−{a}\right){tan}\frac{{x}}{\mathrm{2}}−\sqrt{\mathrm{1}+{a}}}{\left(\mathrm{1}−{a}\right){tan}\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+{a}}}\right)+{c} \\ $$$$ \\ $$

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