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Question-200309

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Question Number 200309 by sonukgindia last updated on 17/Nov/23
Answered by Mathspace last updated on 17/Nov/23
J=∫_0 ^(2π) (dx/(a^2 −2asinx +1)) =∫_0 ^(2π) (dx/(a^2 −2a((e^(ix) −e^(−ix) )/(2i))+1))(e^(ix) =z) =∫_(∣z∣=1) (dz/(iz(a^2 +ia(z−z^(−1) )+1))) =∫_(∣z∣=1) ((−idz)/(a^2 z+iaz^2 −ia +z)) =∫_(∣z∣=1) ((−idz)/(iaz^2 +(a^2 +1)z−ia)) f(z)=((−i)/(iaz^2 +(a^2 +1)z−ia)) poles of f? Δ=(a^2 +1)^2 −4(ia)(−ia) =a^4 +2a^2 +1−4a^2 =(a^2 −1)^2 and (√Δ)=∣a^2 −1∣=1−a^2 z_1 =((−a^2 −1+1−a^2 )/(2ia))=−(1/i)a =ia ⇒∣z_1 ∣=∣a∣<1 z_2 =((−a^2 −1−1+a^2 )/(2ia))=−(1/(ia)) =(i/a) ⇒∣z_2 ∣=(1/a)>1 residus theorem give ∫_(∣z∣=1) f(z)dz=2iπ Res(f,z_1 ) Res(f,z_1 )=lim_(z→z_1 ) (z−z_1 )((−ia)/(ia(z−z_1 )(z−z_2 ))) =((−1)/(z_1 −z_2 ))=((−1)/(ia−(i/a)))=(i/(a^2 −1)) ⇒∫_(∣z∣=1) f(z)dz=2iπ.(i/(a^2 −1)) =((−2π)/(a^2 −1))=((2π)/(1−a^2 )) ⇒ •J=((2π)/(1−a^2 ))•
$${J}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{asinx}\:+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{a}\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}+\mathrm{1}}\left({e}^{{ix}} ={z}\right) \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \frac{{dz}}{{iz}\left({a}^{\mathrm{2}} +{ia}\left({z}−{z}^{−\mathrm{1}} \right)+\mathrm{1}\right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \frac{−{idz}}{{a}^{\mathrm{2}} {z}+{iaz}^{\mathrm{2}} −{ia}\:+{z}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \frac{−{idz}}{{iaz}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +\mathrm{1}\right){z}−{ia}} \\ $$$${f}\left({z}\right)=\frac{−{i}}{{iaz}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +\mathrm{1}\right){z}−{ia}} \\ $$$${poles}\:{of}\:{f}? \\ $$$$\Delta=\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left({ia}\right)\left(−{ia}\right) \\ $$$$={a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} =\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$${and}\:\sqrt{\Delta}=\mid{a}^{\mathrm{2}} −\mathrm{1}\mid=\mathrm{1}−{a}^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\frac{−{a}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{2}{ia}}=−\frac{\mathrm{1}}{{i}}{a} \\ $$$$={ia}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid=\mid{a}\mid<\mathrm{1}\: \\ $$$${z}_{\mathrm{2}} =\frac{−{a}^{\mathrm{2}} −\mathrm{1}−\mathrm{1}+{a}^{\mathrm{2}} }{\mathrm{2}{ia}}=−\frac{\mathrm{1}}{{ia}} \\ $$$$=\frac{{i}}{{a}}\:\Rightarrow\mid{z}_{\mathrm{2}} \mid=\frac{\mathrm{1}}{{a}}>\mathrm{1} \\ $$$${residus}\:{theorem}\:{give} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left({f},{z}_{\mathrm{1}} \right) \\ $$$${Res}\left({f},{z}_{\mathrm{1}} \right)={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\left({z}−{z}_{\mathrm{1}} \right)\frac{−{ia}}{{ia}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$$=\frac{−\mathrm{1}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }=\frac{−\mathrm{1}}{{ia}−\frac{{i}}{{a}}}=\frac{{i}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi.\frac{{i}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{−\mathrm{2}\pi}{{a}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{2}\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\:\:\Rightarrow \\ $$$$\bullet{J}=\frac{\mathrm{2}\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\bullet \\ $$$$ \\ $$
Answered by Frix last updated on 17/Nov/23
∫_0 ^(2π) (dx/(a^2 −2asin x +1))= =2∫_0 ^π (dx/(a^2 −2acos x +1)) =^(t=((1+a)/(1−a))tan (x/2)) =(4/(1−a^2 )) ∫_0 ^∞ (dt/(t^2 +1)) =((2π)/(1−a^2 ))
$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{a}\mathrm{sin}\:{x}\:+\mathrm{1}}= \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{a}\mathrm{cos}\:{x}\:+\mathrm{1}}\:\overset{{t}=\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}\mathrm{tan}\:\frac{{x}}{\mathrm{2}}} {=} \\ $$$$=\frac{\mathrm{4}}{\mathrm{1}−{a}^{\mathrm{2}} }\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\:=\frac{\mathrm{2}\pi}{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$

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