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Question Number 145514 by mathmax by abdo last updated on 05/Jul/21

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}={\int }_0^{2\mathrm{n}\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2}$$$$\mathrm{explicit}{\mathrm{A}}_{\mathrm{n}}\mathrm{and}\mathrm{determine}\mathrm{nature}\mathrm{of}\mathrm{serie}\mathrm{\Sigma }{\mathrm{A}}_{\mathrm{n}}$$

Answered by mathmax by abdo last updated on 05/Jul/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{2\mathrm{n}\pi }\frac{\mathrm{dx}}{\mathrm{a}+\mathrm{cosx}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{2\mathrm{n}\pi }\frac{\mathrm{dx}}{{(\mathrm{a}+\mathrm{cosx})}^2}\Rightarrow$$$${\int }_0^{2\mathrm{n}\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2}=-{\mathrm{f}}^{{}^{\prime }}\left(2\right)\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\int }_{2\mathrm{k}\pi }^{2(\mathrm{k}+1)\pi }\frac{\mathrm{dx}}{\mathrm{a}+\mathrm{cosx}}$$$${=}_{\mathrm{x}=2\mathrm{k}\pi +\mathrm{t}}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\int }_0^{2\pi }\frac{\mathrm{dt}}{\mathrm{a}+\mathrm{cost}}=\mathrm{n}{\int }_0^{2\pi }\frac{\mathrm{dt}}{\mathrm{a}+\mathrm{cost}}$$$$\mathrm{we}\mathrm{have}{\int }_0^{2\pi }\frac{\mathrm{dt}}{\mathrm{a}+\mathrm{cost}}{=}_{{\mathrm{e}}^{\mathrm{it}}=\mathrm{z}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(\mathrm{a}+\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2})}$$$$={\int }_{\mid \mathrm{z}\mid =1}\frac{-2\mathrm{idz}}{(2\mathrm{a}+\mathrm{z}+{\mathrm{z}}^{-1})\mathrm{z}}={\int }_{\mid \mathrm{z}\mid =1}\frac{-2\mathrm{idz}}{{\mathrm{z}}^2+2\mathrm{az}+1}={\int }_{\mid \mathrm{z}\mid =1}\phi \left(\mathrm{z}\right)\mathrm{dz}$$$${\mathrm{\Delta }}^{{}^{\prime }}={\mathrm{a}}^2-1>0(\mathrm{a}>1)\Rightarrow {\mathrm{z}}_1=-\mathrm{a}+\sqrt{{\mathrm{a}}^2-1}$$$$\mathrm{and}{\mathrm{z}}_2=-\mathrm{a}-\sqrt{{\mathrm{a}}^2-1}$$$$\mid {\mathrm{z}}_1\mid -1=\sqrt{{\mathrm{a}}^2-1}-\mathrm{a}-1=\sqrt{{\mathrm{a}}^2-1}-(\mathrm{a}+1)<0\Rightarrow \mid {\mathrm{z}}_1\mid <1$$$$\mid {\mathrm{z}}_2\mid -1=\mathrm{a}+\sqrt{{\mathrm{a}}^2-1}-1=\mathrm{a}-1+\sqrt{{\mathrm{a}}^2-1}>0\Rightarrow \mid {\mathrm{z}}_2\mid >1$$$$\mathrm{residus}\mathrm{theorem}\mathrm{give}{\int }_{\mid \mathrm{z}\mid =1}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,{\mathrm{z}}_1)\mathrm{we}\mathrm{have}$$$$\phi \left(\mathrm{z}\right)=\frac{-2\mathrm{i}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}\Rightarrow \mathrm{Res}(\phi ,{\mathrm{z}}_1)=\frac{-2\mathrm{i}}{{\mathrm{z}}_1-{\mathrm{z}}_2}=\frac{-2\mathrm{i}}{2\sqrt{{\mathrm{a}}^2-1}}\Rightarrow$$$${\int }_{\mid \mathrm{z}\mid =1}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{-\mathrm{i}}{\sqrt{{\mathrm{a}}^2-1}}=\frac{2\pi }{\sqrt{{\mathrm{a}}^2-1}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{2\mathrm{n}\pi }{\sqrt{{\mathrm{a}}^2-1}}=2\mathrm{n}\pi {({\mathrm{a}}^2-1)}^{-\frac{1}{2}}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=2\mathrm{n}\pi (-\frac{1}{2})\left(2\mathrm{a}\right){({\mathrm{a}}^2-1)}^{-\frac{3}{2}}$$$$=-2\mathrm{n}\pi \mathrm{a}{({\mathrm{a}}^2-1)}^{-\frac{3}{2}}=-\frac{2\mathrm{n}\pi \mathrm{a}}{({\mathrm{a}}^2-1)\sqrt{{\mathrm{a}}^2-1}}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(2\right)=\frac{-4\mathrm{n}\pi }{3\sqrt{3}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=-{\mathrm{f}}^{{}^{\prime }}\left(2\right)=\frac{4\mathrm{n}\pi }{3\sqrt{3}}$$$${\mathrm{A}}_{\mathrm{n}}\to +\mathrm{\infty }\Rightarrow \mathrm{\Sigma }{\mathrm{A}}_{\mathrm{n}}\mathrm{is}\mathrm{divergent}...!$$

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