Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 50427 by Abdo msup. last updated on 16/Dec/18

let I_n (λ) =∫_0 ^π ((vos(nt))/(1−2λcost +λ^2 ))dt 1)calculate I_0 (λ) and I_1 (λ) 2) find relation between I_(n−1) ,I_n and I_(n+1) 3) calculate I_n (λ).

$${let}\:{I}_{{n}} \left(\lambda\right)\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{vos}\left({nt}\right)}{\mathrm{1}−\mathrm{2}\lambda{cost}\:+\lambda^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right){calculate}\:{I}_{\mathrm{0}} \left(\lambda\right)\:{and}\:{I}_{\mathrm{1}} \left(\lambda\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{relation}\:{between}\:{I}_{{n}−\mathrm{1}} ,{I}_{{n}} \:{and}\:{I}_{{n}+\mathrm{1}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{I}_{{n}} \left(\lambda\right). \\ $$

Commented by Abdo msup. last updated on 20/Dec/18

1)I_0 (λ)=∫_0 ^π (dt/(1−2λcost +λ^2 )) changement tan((t/2))=x give I_0 (λ)=∫_0 ^∞ (1/(1−2λ((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) =∫_0 ^∞ ((2dx)/(1+x^2 −2λ(1−x^2 ))) =∫_0 ^∞ ((2dx)/((1+2λ)x^2 +1−2λ)) =(2/(1+2λ))∫_0 ^∞ (dx/(x^2 +((1−2λ)/(1+2λ)))) case 1 ((1−2λ)/(1+2λ))>0 we do the changement x=(√((1−2λ)/(1+2λ)))u ⇒I_0 (λ)=(2/(1+2λ))∫_0 ^∞ ((√((1−2λ)/(1+2λ)))/(((1−2λ)/(1+2λ))(1+u^2 )))du =(2/(1+2λ)) ((√(1+2λ))/(√(1−2λ))) ∫_0 ^∞ (du/(1+u^2 )) =(2/(√(1−4λ^2 ))) (π/2) =(π/(√(1−4λ^2 ))) . case 2 ((1−2λ)/(1+2λ))<0 ⇒ I_0 (λ)=(2/(1+2λ)) ∫_0 ^∞ (dx/(x^2 −((√((2λ−1)/(2λ+1))))^2 )) =(2/(1+2λ)) (1/(2(√((2λ−1)/(2λ+1)))))∫_0 ^∞ ((1/(x−(√((2λ−1)/(2λ+1))))) −(1/(x+(√((2λ−1)/(2λ+1))))))dx = (1/(√(4λ^2 −1))) [ln∣((x−(√((2λ−1)/(2λ+1))))/(x+(√((2λ−1)/(2λ+1)))))∣]_0 ^(+∞) =0 .

$$\left.\mathrm{1}\right){I}_{\mathrm{0}} \left(\lambda\right)=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dt}}{\mathrm{1}−\mathrm{2}\lambda{cost}\:+\lambda^{\mathrm{2}} }\:\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={x} \\ $$$${give}\:{I}_{\mathrm{0}} \left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\lambda\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}\lambda\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dx}}{\left(\mathrm{1}+\mathrm{2}\lambda\right){x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}\lambda}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}} \\ $$$${case}\:\mathrm{1}\:\:\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}>\mathrm{0}\:\:{we}\:{do}\:{the}\:{changement}\: \\ $$$${x}=\sqrt{\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}}{u}\:\Rightarrow{I}_{\mathrm{0}} \left(\lambda\right)=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}\lambda}\int_{\mathrm{0}} ^{\infty} \:\frac{\sqrt{\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}}}{\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}\lambda}\:\frac{\sqrt{\mathrm{1}+\mathrm{2}\lambda}}{\sqrt{\mathrm{1}−\mathrm{2}\lambda}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }}\:\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{\pi}{\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }}\:. \\ $$$${case}\:\mathrm{2}\:\:\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\lambda}<\mathrm{0}\:\Rightarrow\:{I}_{\mathrm{0}} \left(\lambda\right)=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}\lambda}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −\left(\sqrt{\frac{\mathrm{2}\lambda−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}\lambda}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{2}\lambda−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{x}−\sqrt{\frac{\mathrm{2}\lambda−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}}}\:−\frac{\mathrm{1}}{{x}+\sqrt{\frac{\mathrm{2}\lambda−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}}}\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{4}\lambda^{\mathrm{2}} −\mathrm{1}}}\:\left[{ln}\mid\frac{{x}−\sqrt{\frac{\mathrm{2}\lambda−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}}}{{x}+\sqrt{\frac{\mathrm{2}\lambda−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}}}\mid\right]_{\mathrm{0}} ^{+\infty} =\mathrm{0}\:. \\ $$

Commented by Abdo msup. last updated on 23/Dec/18

I_1 (λ)=∫_0 ^π ((cos(t))/(1−2λcost +λ^2 ))dt changement tan((t/2))=x give I_1 (λ)=∫_0 ^∞ (((1−x^2 )/(1+x^2 ))/(1−2λ((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) =∫_0 ^∞ ((1−x^2 )/((1+x^2 )^2 ( 1−2λ ((1−x^2 )/(1+x^2 )))))dx =∫_0 ^∞ ((1−x^2 )/((1+x^2 )^(2 ) −2λ(1−x^4 )))dx =∫_0 ^∞ ((1−x^2 )/(x^4 +2x^2 +1 −2λ +2λx^4 ))dx =∫_0 ^∞ ((1−x^2 )/((2λ+1)x^4 +2x^2 +1−2λ))dx let decompose F(x)=((1−x^2 )/((2λ+1)x^4 +2x^2 +1−2λ)) F(x) =g(t) with x^2 =t and g(t) =((1−t^2 )/((2λ+1)t^2 +2t +1−2λ)) Δ^, =1−(1+2λ)(1−2λ)=1−(1−4λ^2 ) =4λ^2 ⇒t_1 = ((−1 +2∣λ∣)/(2λ+1 )) and t_2 =((−1−2∣λ∣)/(2λ +1)) case1 λ>0 ⇒ t_1 =((−1+2λ)/(2λ +1)) and t_2 =((−1−2λ)/(2λ+1))=−1 ⇒ g(t) =((1−t^2 )/((2λ+1)(t−t_1 )(t−t_2 ))) g(t)=−((t^2 −1)/((2λ+1)t^2 +2t +1−2λ)) =−(1/((2λ+1))) (((2λ+1)t^2 −(2λ+1))/((2λ+1)t^2 +2t +1−2λ)) =((−1)/(2λ+1)) { (((2λ+1)t^2 +2t +1−2λ −2t−1+2λ−2λ−1)/((2λ+1)t^2 +2t +1−2λ))} =−(1/(2λ +1)){1 −((2t+2)/((2λ+1)t^2 +2t+ 1−2λ))} =−(1/(2λ +1)) −((2t +2)/((2λ+1){(2λ+1)t^2 +2t +1−2λ})) ....be continued...

$${I}_{\mathrm{1}} \left(\lambda\right)=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{cos}\left({t}\right)}{\mathrm{1}−\mathrm{2}\lambda{cost}\:+\lambda^{\mathrm{2}} }{dt}\:\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}\: \\ $$$${give}\:{I}_{\mathrm{1}} \left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{2}\lambda\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\:\mathrm{1}−\mathrm{2}\lambda\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}\:} −\mathrm{2}\lambda\left(\mathrm{1}−{x}^{\mathrm{4}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}\:−\mathrm{2}\lambda\:+\mathrm{2}\lambda{x}^{\mathrm{4}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda}{dx}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{2}\lambda+\mathrm{1}\right){x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$${F}\left({x}\right)\:={g}\left({t}\right)\:{with}\:{x}^{\mathrm{2}} ={t}\:{and}\:{g}\left({t}\right)\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$$\Delta^{,} \:=\mathrm{1}−\left(\mathrm{1}+\mathrm{2}\lambda\right)\left(\mathrm{1}−\mathrm{2}\lambda\right)=\mathrm{1}−\left(\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} \right) \\ $$$$=\mathrm{4}\lambda^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\:\frac{−\mathrm{1}\:+\mathrm{2}\mid\lambda\mid}{\mathrm{2}\lambda+\mathrm{1}\:}\:\:{and} \\ $$$${t}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{2}\mid\lambda\mid}{\mathrm{2}\lambda\:+\mathrm{1}} \\ $$$${case}\mathrm{1}\:\lambda>\mathrm{0}\:\Rightarrow\:{t}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{2}\lambda}{\mathrm{2}\lambda\:+\mathrm{1}}\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{2}\lambda}{\mathrm{2}\lambda+\mathrm{1}}=−\mathrm{1}\:\Rightarrow \\ $$$${g}\left({t}\right)\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{2}\lambda+\mathrm{1}\right)\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)} \\ $$$${g}\left({t}\right)=−\frac{{t}^{\mathrm{2}} \:−\mathrm{1}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}\lambda+\mathrm{1}\right)}\:\frac{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} −\left(\mathrm{2}\lambda+\mathrm{1}\right)}{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}−\mathrm{2}\lambda} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}\lambda+\mathrm{1}}\:\left\{\:\:\frac{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}−\mathrm{2}\lambda\:−\mathrm{2}{t}−\mathrm{1}+\mathrm{2}\lambda−\mathrm{2}\lambda−\mathrm{1}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}−\mathrm{2}\lambda}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\lambda\:+\mathrm{1}}\left\{\mathrm{1}\:−\frac{\mathrm{2}{t}+\mathrm{2}}{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{2}{t}+\:\mathrm{1}−\mathrm{2}\lambda}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\lambda\:+\mathrm{1}}\:−\frac{\mathrm{2}{t}\:+\mathrm{2}}{\left(\mathrm{2}\lambda+\mathrm{1}\right)\left\{\left(\mathrm{2}\lambda+\mathrm{1}\right){t}^{\mathrm{2}} +\mathrm{2}{t}\:+\mathrm{1}−\mathrm{2}\lambda\right\}} \\ $$$$....{be}\:{continued}... \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com