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Question Number 36946 by maxmathsup by imad last updated on 07/Jun/18
calculateϕ(λ)= ∫_0 ^π ((cos(t))/(1−2λ cost +λ^2 )) dt
$${calculate}\varphi\left(\lambda\right)=\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\:\frac{{cos}\left({t}\right)}{\mathrm{1}−\mathrm{2}\lambda\:{cost}\:+\lambda^{\mathrm{2}} }\:{dt} \\ $$
Commented by math khazana by abdo last updated on 08/Jun/18
changement tan((t/2)) =x give ϕ(λ) = ∫_0 ^∞ (((1−x^2 )/(1+x^2 ))/(1−2λ((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) = 2 ∫_0 ^∞ ((1−x^2 )/((1+x^2 ){1+x^2 −2λ +2λx^2 }))dx =2 ∫_0 ^∞ ((1−x^2 )/((1+x^2 )((2λ+1)x^2 +1−2λ)))dx let decompose F(x)=((1−x^2 )/((1+x^2 ){(2λ+1)x^2 +1−2λ))) F(x)= ((ax+b)/(x^2 +1)) +((cx+d)/((2λ+1)x^2 +1−2λ)) F(−x)=F(x)⇒((−ax+b)/(x^2 +1)) +((−cx +d)/((2λ+1)x^2 +1−2λ)) =F(x) ⇒a=c=0 ⇒F(x)=(b/(x^2 +1)) + (d/((2λ+1)x^2 +1−2λ)) lim_(x→+∞) x^2 F(x)=((−1)/(2λ+1)) = b +(d/(2λ+1)) ⇒ −1=(2λ+1)b +d ⇒d=−1−(2λ+1)b F(x)= (b/(x^2 +1)) −((1+(2λ+1)b)/((2λ+1)x^2 +1−2λ)) F(o)= (1/(1−2λ)) = b −((1+(2λ+1)b)/(1−2λ)) ⇒ 1=(1−2λ)b −1−(2λ+1)b =−4λb −1 ⇒ 2=−4λb ⇒b=((−1)/(2λ)) ( λ≠0 and λ≠(1/2)) ∫_0 ^(+∞) F(x)dx =((−1)/(2λ))∫_0 ^∞ {(1/(x^2 +1)) −((1−((2λ+1)/(2λ)))/((2λ+1)x^2 +1−2λ))}dx =((−1)/(2λ)) ∫_0 ^∞ (dx/(1+x^2 )) −(1/(2λ)) .((−1)/(2λ))∫_0 ^∞ (dx/((2λ+1)x^2 +1−2λ)) =−(1/(2λ)) (π/2) +(1/(4λ^2 )) ∫_0 ^∞ (dx/((2λ+1)x^2 +1−2λ)) but ∫_0 ^∞ (dx/((2λ+1)x^2 +1−2λ)) =(1/(2λ+1))∫_0 ^∞ (dx/(x^2 +((1−2λ)/(1+2λ)))) if ((1−2λ)/(1+2λ)) >0 we use the chang.x=(√((1−2λ)/(1+2λ)))u =(1/(2λ+1))∫_0 ^∞ (1/(((1−2λ)/(1+2λ))(1+x^2 ))) ((√(1−2λ))/( (√(1+2λ)))) du =(1/(2λ+1)) ((1+2λ)/(1−2λ)) ((√(1−2λ))/( (√(1+2λ)))) (π/2) = (π/(2(2λ+1))) (√((1+2λ)/(1−2λ)))=(π/2) (1/( (√(1−4λ^2 )))) ⇒ ∫_0 ^∞ F(x)dx= −(π/(4λ)) +(1/(4λ^2 )) (π/2) (1/( (√(1−4λ^2 )))) =−(π/(4λ)) + (π/(8λ^2 (√(1−4λ^2 )))) ϕ(λ) =2 ∫_0 ^∞ F(x)dx ⇒ ϕ(λ) = (π/(4λ^2 (√(1−4λ^2 )))) −(π/(2λ)) .
$${changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={x}\:{give} \\ $$$$\varphi\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{2}\lambda\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left\{\mathrm{1}+{x}^{\mathrm{2}} \:−\mathrm{2}\lambda\:+\mathrm{2}\lambda{x}^{\mathrm{2}} \right\}}{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\lambda\right)}{dx}\:{let} \\ $$$${decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left\{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda\right)} \\ $$$${F}\left({x}\right)=\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{{cx}+{d}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$${F}\left(−{x}\right)={F}\left({x}\right)\Rightarrow\frac{−{ax}+{b}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cx}\:+{d}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda}\:={F}\left({x}\right) \\ $$$$\Rightarrow{a}={c}=\mathrm{0}\:\Rightarrow{F}\left({x}\right)=\frac{{b}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\:\frac{{d}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$${lim}_{{x}\rightarrow+\infty} {x}^{\mathrm{2}} {F}\left({x}\right)=\frac{−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}\:=\:{b}\:+\frac{{d}}{\mathrm{2}\lambda+\mathrm{1}}\:\Rightarrow \\ $$$$−\mathrm{1}=\left(\mathrm{2}\lambda+\mathrm{1}\right){b}\:+{d}\:\Rightarrow{d}=−\mathrm{1}−\left(\mathrm{2}\lambda+\mathrm{1}\right){b} \\ $$$${F}\left({x}\right)=\:\frac{{b}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{1}+\left(\mathrm{2}\lambda+\mathrm{1}\right){b}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$${F}\left({o}\right)=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\lambda}\:=\:{b}\:−\frac{\mathrm{1}+\left(\mathrm{2}\lambda+\mathrm{1}\right){b}}{\mathrm{1}−\mathrm{2}\lambda}\:\Rightarrow \\ $$$$\mathrm{1}=\left(\mathrm{1}−\mathrm{2}\lambda\right){b}\:−\mathrm{1}−\left(\mathrm{2}\lambda+\mathrm{1}\right){b}\:=−\mathrm{4}\lambda{b}\:−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}=−\mathrm{4}\lambda{b}\:\Rightarrow{b}=\frac{−\mathrm{1}}{\mathrm{2}\lambda}\:\:\left(\:\lambda\neq\mathrm{0}\:{and}\:\lambda\neq\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{+\infty} {F}\left({x}\right){dx}\:=\frac{−\mathrm{1}}{\mathrm{2}\lambda}\int_{\mathrm{0}} ^{\infty} \:\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{1}−\frac{\mathrm{2}\lambda+\mathrm{1}}{\mathrm{2}\lambda}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda}\right\}{dx} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}\lambda}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{2}\lambda}\:.\frac{−\mathrm{1}}{\mathrm{2}\lambda}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\lambda}\:\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}} \\ $$$${if}\:\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}\:>\mathrm{0}\:{we}\:{use}\:{the}\:{chang}.{x}=\sqrt{\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}}{u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\frac{\sqrt{\mathrm{1}−\mathrm{2}\lambda}}{\:\sqrt{\mathrm{1}+\mathrm{2}\lambda}}\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}\:\:\frac{\mathrm{1}+\mathrm{2}\lambda}{\mathrm{1}−\mathrm{2}\lambda}\:\frac{\sqrt{\mathrm{1}−\mathrm{2}\lambda}}{\:\sqrt{\mathrm{1}+\mathrm{2}\lambda}}\:\frac{\pi}{\mathrm{2}} \\ $$$$=\:\frac{\pi}{\mathrm{2}\left(\mathrm{2}\lambda+\mathrm{1}\right)}\:\sqrt{\frac{\mathrm{1}+\mathrm{2}\lambda}{\mathrm{1}−\mathrm{2}\lambda}}=\frac{\pi}{\mathrm{2}}\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{F}\left({x}\right){dx}=\:−\frac{\pi}{\mathrm{4}\lambda}\:+\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }} \\ $$$$=−\frac{\pi}{\mathrm{4}\lambda}\:+\:\frac{\pi}{\mathrm{8}\lambda^{\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }} \\ $$$$\varphi\left(\lambda\right)\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} {F}\left({x}\right){dx}\:\Rightarrow \\ $$$$\varphi\left(\lambda\right)\:=\:\:\frac{\pi}{\mathrm{4}\lambda^{\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }}\:−\frac{\pi}{\mathrm{2}\lambda}\:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 08/Jun/18
if ((1−2λ)/(1+2λ))<0 we get ∫_0 ^∞ (dx/(x^2 +((1−2λ)/(1+2λ)))) =∫_0 ^∞ (dx/(x^2 −((√(−((1−2λ)/(1+2λ)))))^2 )) and we use the decompositon ...
$${if}\:\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}<\mathrm{0}\:\:{we}\:{get}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:\:−\left(\sqrt{−\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}}\right)^{\mathrm{2}} }\:\:\:{and}\:{we}\:{use}\:{the} \\ $$$${decompositon}\:… \\ $$

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