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Question Number 65769 by mathmax by abdo last updated on 03/Aug/19

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2-2xcos\theta +1)}^2}$$

Commented by ~ À ® @ 237 ~ last updated on 04/Aug/19

$$LetsnamedIthevaluewearesearching.Andf\left(x\right)=\frac{1}{{(x^2-2xcos\theta +1)}^2}$$$$x^2-2xcos\theta +1=(x-e^{i\theta })(x-e^{-i\theta })$$$$Im\left(e^{i\theta }\right)=sin\theta >0\Rightarrow \theta \in D=\underset{k=0}{\cup }]2k\pi (2k+1)\pi [etIm\left(e^{-i\theta }\right)<0$$$$Im\left(e^{-i\theta }\right)=-sin\theta >0\Rightarrow \theta \notin D$$$$if\theta \in Dwewillhavesin\theta >0$$$$SoI=2i\pi Res(f.e^{i\theta })$$$$AndRes(f.e^{i\theta })={lim}_{z->e^{i\theta }}\frac{1}{(2-1)!}{\left[\frac{1}{{(z-e^{-i\theta })}^2}\right]}^{\left(1\right)}=\frac{-2}{{(e^{i\theta }-e^{-i\theta })}^3}=\frac{-i}{4{\left(sin\theta \right)}^3}$$$$ThenI=\frac{\pi }{2{sin}^3\left(\theta \right)}$$$$if\theta \notin Dthensin\theta <0$$$$soI=2i\pi Res(f.e^{-i\theta })$$$$andRes(f.e^{-i\theta })={lim}_{z->e^{-i\theta }}\frac{1}{(2-1)!}{\left[\frac{1}{{(z-e^{i\theta })}^2}\right]}^{\left(1\right)}=\frac{-2}{{(e^{-i\theta }-e^{i\theta })}^3}=\frac{i}{4{sin}^3\theta }$$$$thenI=\frac{-\pi }{2{sin}^3\left(\theta \right)}$$$$FinallywecanconcludethatI=\frac{\pi }{2\mid sin\theta {\mid }^3}$$

Commented by mathmax by abdo last updated on 04/Aug/19

$$thankyousir$$

Commented by mathmax by abdo last updated on 04/Aug/19

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{x^2-2xcos\theta +a}witha>{cos}^2\theta$$$$wehave{f}^{{}^{\prime }}\left(a\right)=-{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2-2xcos\theta +a)}^2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2-2xcos\theta +1)}^2}=-f^{{}^{\prime }}\left(1\right)letfirstfindf\left(a\right)$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{x^2-2xcos\theta +{cos}^2\theta +a-{cos}^2\theta }$$$$={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x-cos\theta )}^2+a-{cos}^2\theta }changementx-cos\theta =\sqrt{a-{cos}^2\theta }u$$$$givef\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{\sqrt{a-{cos}^2\theta }du}{(a-{cos}^2\theta )(1+u^2)}=\frac{1}{\sqrt{a-{cos}^2\theta }}\frac{\pi }{2}=\frac{\pi }{2\sqrt{a-{cos}^2\theta }}$$$${\Rightarrow f^{{}^{\prime }}\left(a\right)=\frac{\pi }{2}{(a-{cos}^2\theta )}^{-\frac{1}{2}}\}}^{\left(1\right)}=-\frac{\pi }{4}{(a-{cos}^2\theta )}^{-\frac{3}{2}}$$$$=-\frac{\pi }{4(a-{cos}^2\theta )\sqrt{a-{cos}^2\theta }}\Rightarrow f^{{}^{\prime }}\left(1\right)=\frac{\pi }{4(1-{cos}^2\theta )\sqrt{1-{cos}^2\theta }}$$$$=\frac{\pi }{4{sin}^2\theta \mid sin\theta \mid }\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2-2xcos\theta +1)}^2}=\frac{\pi }{4\mid sin\theta {\mid }^3}$$

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