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

Answered by Mathspace last updated on 16/Nov/23

$$I={\int }_0^{2\pi }\frac{dx}{a^2-2acosx+1}dx$$$$={\int }_0^{2\pi }\frac{dx}{a^2-2a\frac{e^{ix}+e^{-ix}}{2}+1}(z=e^{ix})$$$$={\int }_{\mid z\mid =1}\frac{dz}{iz(a^2-a(z+z^{-1})+1)}$$$$={\int }_{\mid z\mid =1}\frac{-idz}{a^{2}z-{az}^2-a+z}$$$$={\int }_{\mid z\mid =1}\frac{-idz}{-{az}^2+(a^2+1)z-a}$$$$={\int }_{\mid z\mid =1}\frac{idz}{{az}^2-(a^2+1)z+a}$$$$letf\left(z\right)=\frac{i}{{az}^2-(a^2+1)z+a}$$$$polesoff?$$$$\mathrm{\Delta }={(a^2+1)}^2-4a^2$$$$=a^4+2a^2+1-4a^2=a^4-2a^2+1$$$$={(a^2-1)}^{2}and\sqrt{\mathrm{\Delta }}=\mid a^2-1\mid$$$$=1-a^{2}duotoo<a<1$$$$\Rightarrow z_1=\frac{a^2+1+1-a^2}{2a}=\frac{1}{a}$$$$o<a<1\Rightarrow \frac{1}{a}>1\Rightarrow \mid z_1\mid >1$$$$z_2=\frac{a^2+1-1+a^2}{2a}=aand$$$$\mid z_2\mid =a<1residustheoremgive$$$${\int }_{\mid z\mid }f\left(z\right)dz=2i\pi Res(f,z_2)$$$$wehavef\left(z\right)=\frac{i}{a(z-z_1)(z-z_2)}$$$$\Rightarrow Res(f,z_2)=\frac{i}{a(z_2-z_1)}$$$$=\frac{i}{a.(a-\frac{1}{a})}=\frac{i}{a^2-1}$$$$\Rightarrow {\int }_{\mid z\mid =1}f\left(z\right)dz=2i\pi \times \frac{i}{a^2-1}$$$$=\frac{-2\pi }{a^2-1}=\frac{2\pi }{1-a^2}andfinally$$$$★I=\frac{2\pi }{1-a^2}★$$$$$$

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