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Question Number 57746 by maxmathsup by imad last updated on 10/Apr/19
let f(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2xt +1)^2 )) with ∣x∣<1 (x real) 1) determine a explicit form for f(x) 2) find also g(x) =∫_(−∞) ^(+∞) ((tdt)/((t^2 −2xt +1)^3 )) 3) calculate ∫_(−∞) ^(+∞) (dt/((t^2 −(√2)t +1)^2 )) and ∫_(−∞) ^(+∞) ((tdt)/((t^2 −(√2)t +1)^3 )) 4) calculate A(θ) =∫_(−∞) ^(+∞) (dt/((t^2 −2cosθ t+1)^2 )) and B(θ) =∫_(−∞) ^(+∞) ((tdt)/((t^2 −2cosθ t +1)^3 )) with 0<θ <(π/2) .
$${let}\:{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{with}\:\mid{x}\mid<\mathrm{1}\:\:\:\left({x}\:{real}\right) \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{also}\:{g}\left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{and}\:\int_{−\infty} ^{+\infty} \:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:{A}\left(\theta\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:−\mathrm{2}{cos}\theta\:{t}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{and}\: \\ $$$${B}\left(\theta\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} \:−\mathrm{2}{cos}\theta\:{t}\:+\mathrm{1}\right)^{\mathrm{3}} }\:\:\:\:{with}\:\mathrm{0}<\theta\:<\frac{\pi}{\mathrm{2}}\:\:\:\:\:. \\ $$
Commented by maxmathsup by imad last updated on 13/Apr/19
1) let consider the complex function ϕ(z) =(1/((z^2 −2xz +1)^2 )) let determine the poles of ϕ z^2 −2xz +1 =0 Δ^′ =x^2 −1 =(i(√(1−x^2 )))^2 ⇒z_1 =x+i(√(1−x^2 )) and z_2 =x−i(√(1−x^2 )) ϕ(z) =(1/((z−z_1 )^2 (z−z_2 )^2 )) the poles of ϕ are z_1 and z_2 (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {(1/((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) ((−2(z−z_2 ))/((z−z_2 )^4 )) =lim_(z→z_2 ) ((−2)/((z−z_2 )^3 )) =((−2)/((z_1 −z_2 )^3 )) =((−2)/((2i(√(1−x^2 )))^3 )) =((−2)/(−8i(1−x^2 )(√(1−x^2 )))) =(1/(4i(1−x^2 )(√(1−x^2 )))) ⇒ f(x) =∫_(−∞) ^(+∞) ϕ(z)dz = 2iπ (1/(4i(1−x^2 )(√(1−x^2 )))) ⇒f(x) =(π/(2(1−x^2 )(√(1−x^2 )))) with ∣x∣<1 .
$$\left.\mathrm{1}\right)\:{let}\:{consider}\:{the}\:{complex}\:{function}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −\mathrm{2}{xz}\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{determine}\:{the}\:{poles}\:{of}\:\varphi\:\:\:{z}^{\mathrm{2}} −\mathrm{2}{xz}\:+\mathrm{1}\:=\mathrm{0}\: \\ $$$$\Delta^{'} \:={x}^{\mathrm{2}} −\mathrm{1}\:=\left({i}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} ={x}+{i}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{and}\:{z}_{\mathrm{2}} ={x}−{i}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} \:\left({doubles}\right) \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} \:} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \:\varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\left\{\frac{\mathrm{1}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{−\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right)}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} }\:={lim}_{{z}\rightarrow{z}_{\mathrm{2}} } \:\:\:\:\frac{−\mathrm{2}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{2}}{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{\left(\mathrm{2}{i}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{−\mathrm{8}{i}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\mathrm{4}{i}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{4}{i}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${with}\:\mid{x}\mid<\mathrm{1}\:. \\ $$
Commented by maxmathsup by imad last updated on 13/Apr/19
2) we have f^′ (x) =∫_(−∞) ^(+∞) (∂/∂x)((1/((t^2 −2xt +1)^2 )))dt =∫_(−∞) ^(+∞) ((2t(t^2 −2xt +1))/((t^2 −2xt +1)^4 )) dt = 2 ∫_(−∞) ^(+∞) ((t dt)/((t^2 −2xt +1)^3 )) =2g(x) ⇒g(x)=(1/2)f^′ (x) we have f(x) =(π/(2(1−x^2 )(√(1−x^2 )))) ⇒f(x) =(π/2){(1−x^2 )^(−1) (1−x^2 )^(−(1/2)) } ⇒ f^′ (x) =(π/2){2x (1−x^2 )^(−2) (1−x^2 )^(−(1/2)) +x(1−x^2 )^(−(3/2)) (1−x^2 )^(−1) } =(π/2){ ((2x)/((1−x^2 )^2 (√(1−x^2 )))) +(x/((1−x^2 )(1−x^2 )(√(1−x^2 ))))} =((πx)/2){ (2/((1−x^2 )^2 (√(1−x^2 )))) +(x/((1−x^2 )^2 (√(1−x^2 ))))} =((πx(x+2))/(2(1−x^2 )^2 (√(1−x^2 )))) ⇒g(x) =((πx(x+2))/(4(1−x^2 )^2 (√(1−x^2 )))) .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{\partial}{\partial{x}}\left(\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{1}\right)^{\mathrm{2}} }\right){dt} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{2}{t}\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{1}\right)}{\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{1}\right)^{\mathrm{4}} }\:{dt}\:=\:\mathrm{2}\:\int_{−\infty} ^{+\infty} \:\:\frac{{t}\:{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{1}\right)^{\mathrm{3}} }\:=\mathrm{2}{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{f}^{'} \left({x}\right)\: \\ $$$${we}\:{have}\:{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}\left\{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\frac{\pi}{\mathrm{2}}\left\{\mathrm{2}{x}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\:+{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\:\:\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+\frac{{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right\} \\ $$$$=\frac{\pi{x}}{\mathrm{2}}\left\{\:\:\frac{\mathrm{2}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+\frac{{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right\} \\ $$$$=\frac{\pi{x}\left({x}+\mathrm{2}\right)}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow{g}\left({x}\right)\:=\frac{\pi{x}\left({x}+\mathrm{2}\right)}{\mathrm{4}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 13/Apr/19
3) ∫_(−∞) ^(+∞) (dt/((t^2 −(√2)t +1)^2 )) =f(((√2)/2)) = (π/(2(1−(((√2)/2))^2 (√(1−(((√2)/2))^2 )))) =(π/(2(1−(1/2))(√(1−(1/2))))) =(π/( (√(1/2)))) =π(√2) ∫_(−∞) ^(+∞) ((tdt)/((t^2 −(√2)t +1)^3 )) =g(((√2)/2)) =((π((√2)/2)(((√2)/2) +2))/(4(1−(((√2)/2))^2 )^2 (√(1−(((√2)/2))^2 )))) =((π(√2)(4+(√2)))/(16(1−(1/2))^2 (√(1−(1/2))))) =((π(√2)(4+(√2)))/(4.(1/( (√2))))) =((2π(4+(√2)))/4) =(π/2)(4+(√2)) .
$$\left.\mathrm{3}\right)\:\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)^{\mathrm{2}} }\:={f}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }\right.} \\ $$$$=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}}\:=\frac{\pi}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}\:=\pi\sqrt{\mathrm{2}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)^{\mathrm{3}} }\:={g}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:=\frac{\pi\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\mathrm{2}\right)}{\mathrm{4}\left(\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \right)^{\mathrm{2}} \sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\pi\sqrt{\mathrm{2}}\left(\mathrm{4}+\sqrt{\mathrm{2}}\right)}{\mathrm{16}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}}\:=\frac{\pi\sqrt{\mathrm{2}}\left(\mathrm{4}+\sqrt{\mathrm{2}}\right)}{\mathrm{4}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\:=\frac{\mathrm{2}\pi\left(\mathrm{4}+\sqrt{\mathrm{2}}\right)}{\mathrm{4}}\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{4}+\sqrt{\mathrm{2}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 13/Apr/19
4) we have A(θ) =∫_(−∞) ^(+∞) (dt/((t^2 −2cosθ t +1)^2 )) =f(cosθ)=(π/(2(1−cos^2 θ)(√(1−cos^2 θ)))) =(π/(2sin^2 θ sinθ)) =(π/(2sin^3 θ)) ( 0<θ<(π/2)) also we have B(θ) =∫_(−∞) ^(+∞) ((tdt)/((t^2 −2cosθ t +1)^3 )) ⇒B(θ) =g(cosθ) =((π cosθ(cosθ +2))/(4(1−cos^2 θ)^2 (√(1−cos^2 θ)))) =((π cosθ(2+cosθ))/(4sin^4 θ sinθ)) =((2π cosθ +π cos^2 θ)/(4 sin^5 θ)) .
$$\left.\mathrm{4}\right)\:{we}\:{have}\:{A}\left(\theta\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:−\mathrm{2}{cos}\theta\:{t}\:+\mathrm{1}\right)^{\mathrm{2}} }\:={f}\left({cos}\theta\right)=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \theta}} \\ $$$$=\frac{\pi}{\mathrm{2}{sin}^{\mathrm{2}} \theta\:{sin}\theta}\:=\frac{\pi}{\mathrm{2}{sin}^{\mathrm{3}} \theta}\:\:\left(\:\:\:\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\right) \\ $$$${also}\:{we}\:{have}\:{B}\left(\theta\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{cos}\theta\:{t}\:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow{B}\left(\theta\right)\:={g}\left({cos}\theta\right) \\ $$$$=\frac{\pi\:{cos}\theta\left({cos}\theta\:+\mathrm{2}\right)}{\mathrm{4}\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} \sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \theta}}\:=\frac{\pi\:{cos}\theta\left(\mathrm{2}+{cos}\theta\right)}{\mathrm{4}{sin}^{\mathrm{4}} \theta\:{sin}\theta}\:=\frac{\mathrm{2}\pi\:{cos}\theta\:+\pi\:{cos}^{\mathrm{2}} \theta}{\mathrm{4}\:{sin}^{\mathrm{5}} \theta}\:. \\ $$$$ \\ $$

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