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Question Number 65769 by mathmax by abdo last updated on 03/Aug/19
find the value of ∫_0 ^∞ (dx/((x^2 −2xcosθ +1)^2 ))
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by ~ À ® @ 237 ~ last updated on 04/Aug/19
Lets named I the value we are searching.And f(x)=(1/((x^2 −2xcosθ+1)^2 )) x^2 −2xcosθ+1=(x−e^(iθ) )(x−e^(−iθ) ) Im(e^(iθ) )=sinθ >0 ⇒θ∈D=∪_(k=0) ]2kπ (2k+1)π[ et Im(e^(−iθ) )<0 Im(e^(−iθ) )=−sinθ >0 ⇒ θ∉ D if θ∈D we will have sinθ>0 So I=2iπRes(f . e^(iθ) ) And Res(f. e^(iθ) )= lim_(z−>e^(iθ) ) (1/((2−1)!))[ (1/((z−e^(−iθ) )^2 ))]^((1)) =((−2)/((e^(iθ) −e^(−iθ) )^3 )) =((−i)/(4(sinθ)^3 )) Then I=(π/(2sin^3 (θ))) if θ∉D then sinθ<0 so I= 2iπRes(f. e^(−iθ) ) and Res(f. e^(−iθ) )=lim_(z−>e^(−iθ) ) (1/((2−1)!))[(1/((z−e^(iθ) )^2 ))]^((1)) =((−2)/((e^(−iθ) −e^(iθ) )^3 ))=(i/(4sin^3 θ)) then I=((−π)/(2sin^3 (θ))) Finally we can conclude that I=(π/(2∣sinθ∣^3 ))
$${Lets}\:{named}\:{I}\:{the}\:{value}\:{we}\:{are}\:{searching}.{And}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta+\mathrm{1}=\left({x}−{e}^{{i}\theta} \right)\left({x}−{e}^{−{i}\theta} \right) \\ $$$$\left.{Im}\left({e}^{{i}\theta} \right)={sin}\theta\:>\mathrm{0}\:\Rightarrow\theta\in{D}=\underset{{k}=\mathrm{0}} {\cup}\right]\mathrm{2}{k}\pi\:\:\:\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\left[\:\:{et}\:{Im}\left({e}^{−{i}\theta} \right)<\mathrm{0}\right. \\ $$$${Im}\left({e}^{−{i}\theta} \right)=−{sin}\theta\:>\mathrm{0}\:\Rightarrow\:\theta\notin\:{D} \\ $$$${if}\:\theta\in{D}\:{we}\:{will}\:{have}\:{sin}\theta>\mathrm{0} \\ $$$${So}\:{I}=\mathrm{2}{i}\pi{Res}\left({f}\:.\:{e}^{{i}\theta} \right) \\ $$$${And}\:{Res}\left({f}.\:{e}^{{i}\theta} \right)=\:{lim}_{{z}−>{e}^{{i}\theta} } \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left[\:\frac{\mathrm{1}}{\left({z}−{e}^{−{i}\theta} \right)^{\mathrm{2}} }\right]^{\left(\mathrm{1}\right)} =\frac{−\mathrm{2}}{\left({e}^{{i}\theta} −{e}^{−{i}\theta} \right)^{\mathrm{3}} }\:=\frac{−{i}}{\mathrm{4}\left({sin}\theta\right)^{\mathrm{3}} } \\ $$$${Then}\:\:{I}=\frac{\pi}{\mathrm{2}{sin}^{\mathrm{3}} \left(\theta\right)} \\ $$$${if}\:\theta\notin{D}\:\:\:{then}\:{sin}\theta<\mathrm{0} \\ $$$${so}\:{I}=\:\mathrm{2}{i}\pi{Res}\left({f}.\:\:{e}^{−{i}\theta} \right) \\ $$$${and}\:{Res}\left({f}.\:\:{e}^{−{i}\theta} \right)={lim}_{{z}−>{e}^{−{i}\theta} } \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left[\frac{\mathrm{1}}{\left({z}−{e}^{{i}\theta} \right)^{\mathrm{2}} }\right]^{\left(\mathrm{1}\right)} =\frac{−\mathrm{2}}{\left({e}^{−{i}\theta} −{e}^{{i}\theta} \right)^{\mathrm{3}} }=\frac{{i}}{\mathrm{4}{sin}^{\mathrm{3}} \theta} \\ $$$${then}\:{I}=\frac{−\pi}{\mathrm{2}{sin}^{\mathrm{3}} \left(\theta\right)} \\ $$$${Finally}\:{we}\:{can}\:{conclude}\:{that}\:{I}=\frac{\pi}{\mathrm{2}\mid{sin}\theta\mid^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 04/Aug/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 04/Aug/19
let f(a) =∫_0 ^∞ (dx/(x^2 −2xcosθ +a)) with a>cos^2 θ we have f^′ (a) =−∫_0 ^∞ (dx/((x^2 −2xcosθ +a)^2 )) ⇒ ∫_0 ^∞ (dx/((x^2 −2xcosθ +1)^2 )) =−f^′ (1) let first find f(a) f(a) =∫_0 ^∞ (dx/(x^2 −2xcosθ +cos^2 θ +a−cos^2 θ)) =∫_0 ^∞ (dx/((x−cosθ)^2 +a−cos^2 θ)) changement x−cosθ =(√(a−cos^2 θ))u give f(a) =∫_0 ^∞ (((√(a−cos^2 θ))du)/((a−cos^2 θ)(1+u^2 ))) =(1/( (√(a−cos^2 θ))))(π/2) =(π/(2(√(a−cos^2 θ)))) ⇒f^′ (a) =(π/2)(a−cos^2 θ)^(−(1/2)) }^((1)) =−(π/4)(a−cos^2 θ)^(−(3/2)) =−(π/(4(a−cos^2 θ)(√(a−cos^2 θ)))) ⇒f^′ (1) =(π/(4(1−cos^2 θ)(√(1−cos^2 θ)))) =(π/(4sin^2 θ∣sinθ∣)) ⇒ ∫_0 ^∞ (dx/((x^2 −2xcosθ +1)^2 )) =(π/(4∣sinθ∣^3 ))
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+{a}}\:\:\:\:\:{with}\:{a}>{cos}^{\mathrm{2}} \theta \\ $$$${we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+{a}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}\right)^{\mathrm{2}} }\:=−{f}^{'} \left(\mathrm{1}\right)\:\:{let}\:{first}\:{find}\:{f}\left({a}\right) \\ $$$${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+{cos}^{\mathrm{2}} \theta\:+{a}−{cos}^{\mathrm{2}} \theta} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}−{cos}\theta\right)^{\mathrm{2}} \:+{a}−{cos}^{\mathrm{2}} \theta}\:{changement}\:{x}−{cos}\theta\:=\sqrt{{a}−{cos}^{\mathrm{2}} \theta}{u} \\ $$$${give}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{a}−{cos}^{\mathrm{2}} \theta}{du}}{\left({a}−{cos}^{\mathrm{2}} \theta\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\:\sqrt{{a}−{cos}^{\mathrm{2}} \theta}}\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}\sqrt{{a}−{cos}^{\mathrm{2}} \theta}} \\ $$$$\left.\Rightarrow{f}^{'} \left({a}\right)\:=\frac{\pi}{\mathrm{2}}\left({a}−{cos}^{\mathrm{2}} \theta\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}^{\left(\mathrm{1}\right)} =−\frac{\pi}{\mathrm{4}}\left({a}−{cos}^{\mathrm{2}} \theta\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=−\frac{\pi}{\mathrm{4}\left({a}−{cos}^{\mathrm{2}} \theta\right)\sqrt{{a}−{cos}^{\mathrm{2}} \theta}}\:\Rightarrow{f}^{'} \left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{4}\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \theta}} \\ $$$$=\frac{\pi}{\mathrm{4}{sin}^{\mathrm{2}} \theta\mid{sin}\theta\mid}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{4}\mid{sin}\theta\mid^{\mathrm{3}} } \\ $$

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