Question-200275

Question Number 200275 by sonukgindia last updated on 16/Nov/23

Answered by Mathspace last updated on 16/Nov/23

I=∫_0 ^(2π) (dx/(a^2 −2acosx+1))dx =∫_0 ^(2π) (dx/(a^2 −2a((e^(ix) +e^(−ix) )/2)+1))(z=e^(ix) ) =∫_(∣z∣=1) (dz/(iz(a^2 −a(z+z^(−1) )+1))) =∫_(∣z∣=1) ((−idz)/(a^2 z−az^2 −a+z)) =∫_(∣z∣=1) ((−idz)/(−az^2 +(a^2 +1)z−a)) =∫_(∣z∣=1) ((idz)/(az^2 −(a^2 +1)z+a)) let f(z)=(i/(az^2 −(a^2 +1)z+a)) poles of f? Δ=(a^2 +1)^2 −4a^2 =a^4 +2a^2 +1−4a^2 =a^4 −2a^2 +1 =(a^2 −1)^2 and (√Δ)=∣a^2 −1∣ =1−a^2 duo to o<a<1 ⇒z_1 =((a^2 +1+1−a^2 )/(2a))=(1/a) o<a<1 ⇒(1/a)>1 ⇒∣z_1 ∣>1 z_2 =((a^2 +1−1+a^2 )/(2a))=a and ∣z_2 ∣=a<1 residus theorem give ∫_(∣z∣) f(z)dz=2iπRes(f,z_2 ) we have f(z)=(i/(a(z−z_1 )(z−z_2 ))) ⇒Res(f,z_2 )=(i/(a(z_2 −z_1 ))) =(i/(a.(a−(1/a))))=(i/(a^2 −1)) ⇒∫_(∣z∣=1) f(z)dz=2iπ×(i/(a^2 −1)) =((−2π)/(a^2 −1))=((2π)/(1−a^2 )) and finally ★I=((2π)/(1−a^2 ))★

$${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{acosx}+\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{a}\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}+\mathrm{1}}\left({z}={e}^{{ix}} \right) \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{{dz}}{{iz}\left({a}^{\mathrm{2}} −{a}\left({z}+{z}^{−\mathrm{1}} \right)+\mathrm{1}\right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\frac{−{idz}}{{a}^{\mathrm{2}} {z}−{az}^{\mathrm{2}} −{a}+{z}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−{idz}}{−{az}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +\mathrm{1}\right){z}−{a}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \frac{{idz}}{{az}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +\mathrm{1}\right){z}+{a}} \\ $$$${let}\:{f}\left({z}\right)=\frac{{i}}{{az}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +\mathrm{1}\right){z}+{a}} \\ $$$${poles}\:{of}\:{f}? \\ $$$$\Delta=\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} ={a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1} \\ $$$$=\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:{and}\:\sqrt{\Delta}=\mid{a}^{\mathrm{2}} −\mathrm{1}\mid \\ $$$$=\mathrm{1}−{a}^{\mathrm{2}} \:{duo}\:{to}\:{o}<{a}<\mathrm{1} \\ $$$$\Rightarrow{z}_{\mathrm{1}} =\frac{{a}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{2}{a}}=\frac{\mathrm{1}}{{a}} \\ $$$${o}<{a}<\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{{a}}>\mathrm{1}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid>\mathrm{1} \\ $$$${z}_{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}+{a}^{\mathrm{2}} }{\mathrm{2}{a}}={a}\:{and} \\ $$$$\mid{z}_{\mathrm{2}} \mid={a}<\mathrm{1}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{\mid{z}\mid} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f},{z}_{\mathrm{2}} \right) \\ $$$${we}\:{have}\:{f}\left({z}\right)=\frac{{i}}{{a}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$$\Rightarrow{Res}\left({f},{z}_{\mathrm{2}} \right)=\frac{{i}}{{a}\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)} \\ $$$$=\frac{{i}}{{a}.\left({a}−\frac{\mathrm{1}}{{a}}\right)}=\frac{{i}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi×\frac{{i}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{−\mathrm{2}\pi}{{a}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{2}\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\:{and}\:{finally} \\ $$$$\bigstar{I}=\frac{\mathrm{2}\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\bigstar \\ $$$$ \\ $$

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