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


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

$l e t I_{n} (λ) = \int_{0}^{π} \frac{v o s (n t)}{1 - 2 λ c o s t + λ^{2}} d t$ $1) c a l c u l a t e I_{0} (λ) a n d I_{1} (λ)$ $2) f i n d r e l a t i o n b e t w e e n I_{n - 1}, I_{n} a n d I_{n + 1}$ $3) c a l c u l a t e I_{n} (λ) .$
Commented by Abdo msup. last updated on 20/Dec/18

$1) I_{0} (λ) = \int_{0}^{π} \frac{d t}{1 - 2 λ c o s t + λ^{2}} c h a n g e m e n t t a n (\frac{t}{2}) = x$ $g i v e I_{0} (λ) = \int_{0}^{\infty} \frac{1}{1 - 2 λ \frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}}$ $= \int_{0}^{\infty} \frac{2 d x}{1 + x^{2} - 2 λ (1 - x^{2})} = \int_{0}^{\infty} \frac{2 d x}{(1 + 2 λ) x^{2} + 1 - 2 λ}$ $= \frac{2}{1 + 2 λ} \int_{0}^{\infty} \frac{d x}{x^{2} + \frac{1 - 2 λ}{1 + 2 λ}}$ $c a s e 1 \frac{1 - 2 λ}{1 + 2 λ} > 0 w e d o t h e c h a n g e m e n t$ $x = \sqrt{\frac{1 - 2 λ}{1 + 2 λ}} u \Rightarrow I_{0} (λ) = \frac{2}{1 + 2 λ} \int_{0}^{\infty} \frac{\sqrt{\frac{1 - 2 λ}{1 + 2 λ}}}{\frac{1 - 2 λ}{1 + 2 λ} (1 + u^{2})} d u$ $= \frac{2}{1 + 2 λ} \frac{\sqrt{1 + 2 λ}}{\sqrt{1 - 2 λ}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} = \frac{2}{\sqrt{1 - 4 λ^{2}}} \frac{π}{2}$ $= \frac{π}{\sqrt{1 - 4 λ^{2}}} .$ $c a s e 2 \frac{1 - 2 λ}{1 + 2 λ} < 0 \Rightarrow I_{0} (λ) = \frac{2}{1 + 2 λ} \int_{0}^{\infty} \frac{d x}{x^{2} - {(\sqrt{\frac{2 λ - 1}{2 λ + 1}})}^{2}}$ $= \frac{2}{1 + 2 λ} \frac{1}{2 \sqrt{\frac{2 λ - 1}{2 λ + 1}}} \int_{0}^{\infty} (\frac{1}{x - \sqrt{\frac{2 λ - 1}{2 λ + 1}}} - \frac{1}{x + \sqrt{\frac{2 λ - 1}{2 λ + 1}}}) d x$ $= \frac{1}{\sqrt{4 λ^{2} - 1}} {[l n ∣ \frac{x - \sqrt{\frac{2 λ - 1}{2 λ + 1}}}{x + \sqrt{\frac{2 λ - 1}{2 λ + 1}}} ∣]}_{0}^{+ \infty} = 0 .$
Commented by Abdo msup. last updated on 23/Dec/18
$I_1 (λ)=∫_0 ^π ((cos(t))/(1−2λcost +λ^2 ))dt changement tan((t/2))=x give I_1 (λ)=∫_0 ^∞ (((1−x^2 )/(1+x^2 ))/(1−2λ((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) =∫_0 ^∞ ((1−x^2 )/((1+x^2 )^2 ( 1−2λ ((1−x^2 )/(1+x^2 )))))dx =∫_0 ^∞ ((1−x^2 )/((1+x^2 )^(2 ) −2λ(1−x^4 )))dx =∫_0 ^∞ ((1−x^2 )/(x^4 +2x^2 +1 −2λ +2λx^4 ))dx =∫_0 ^∞ ((1−x^2 )/((2λ+1)x^4 +2x^2 +1−2λ))dx let decompose F(x)=((1−x^2 )/((2λ+1)x^4 +2x^2 +1−2λ)) F(x) =g(t) with x^2 =t and g(t) =((1−t^2 )/((2λ+1)t^2 +2t +1−2λ)) Δ^, =1−(1+2λ)(1−2λ)=1−(1−4λ^2 ) =4λ^2 ⇒t_1 = ((−1 +2∣λ∣)/(2λ+1 )) and t_2 =((−1−2∣λ∣)/(2λ +1)) case1 λ>0 ⇒ t_1 =((−1+2λ)/(2λ +1)) and t_2 =((−1−2λ)/(2λ+1))=−1 ⇒ g(t) =((1−t^2 )/((2λ+1)(t−t_1 )(t−t_2 ))) g(t)=−((t^2 −1)/((2λ+1)t^2 +2t +1−2λ)) =−(1/((2λ+1))) (((2λ+1)t^2 −(2λ+1))/((2λ+1)t^2 +2t +1−2λ)) =((−1)/(2λ+1)) { (((2λ+1)t^2 +2t +1−2λ −2t−1+2λ−2λ−1)/((2λ+1)t^2 +2t +1−2λ))} =−(1/(2λ +1)){1 −((2t+2)/((2λ+1)t^2 +2t+ 1−2λ))} =−(1/(2λ +1)) −((2t +2)/((2λ+1){(2λ+1)t^2 +2t +1−2λ})) ....be continued...$
$I_{1} (λ) = \int_{0}^{π} \frac{c o s (t)}{1 - 2 λ c o s t + λ^{2}} d t c h a n g e m e n t t a n (\frac{t}{2}) = x$ $g i v e I_{1} (λ) = \int_{0}^{\infty} \frac{\frac{1 - x^{2}}{1 + x^{2}}}{1 - 2 λ \frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}}$ $= \int_{0}^{\infty} \frac{1 - x^{2}}{{(1 + x^{2})}^{2} (1 - 2 λ \frac{1 - x^{2}}{1 + x^{2}})} d x$ $= \int_{0}^{\infty} \frac{1 - x^{2}}{{(1 + x^{2})}^{2} - 2 λ (1 - x^{4})} d x$ $= \int_{0}^{\infty} \frac{1 - x^{2}}{x^{4} + 2 x^{2} + 1 - 2 λ + 2 λ x^{4}} d x$ $= \int_{0}^{\infty} \frac{1 - x^{2}}{(2 λ + 1) x^{4} + 2 x^{2} + 1 - 2 λ} d x l e t d e c o m p o s e$ $F (x) = \frac{1 - x^{2}}{(2 λ + 1) x^{4} + 2 x^{2} + 1 - 2 λ}$ $F (x) = g (t) w i t h x^{2} = t a n d g (t) = \frac{1 - t^{2}}{(2 λ + 1) t^{2} + 2 t + 1 - 2 λ}$ $Δ^{,} = 1 - (1 + 2 λ) (1 - 2 λ) = 1 - (1 - 4 λ^{2})$ $= 4 λ^{2} \Rightarrow t_{1} = \frac{- 1 + 2 ∣ λ ∣}{2 λ + 1} a n d$ $t_{2} = \frac{- 1 - 2 ∣ λ ∣}{2 λ + 1}$ $c a s e 1 λ > 0 \Rightarrow t_{1} = \frac{- 1 + 2 λ}{2 λ + 1} a n d t_{2} = \frac{- 1 - 2 λ}{2 λ + 1} = - 1 \Rightarrow$ $g (t) = \frac{1 - t^{2}}{(2 λ + 1) (t - t_{1}) (t - t_{2})}$ $g (t) = - \frac{t^{2} - 1}{(2 λ + 1) t^{2} + 2 t + 1 - 2 λ}$ $= - \frac{1}{(2 λ + 1)} \frac{(2 λ + 1) t^{2} - (2 λ + 1)}{(2 λ + 1) t^{2} + 2 t + 1 - 2 λ}$ $= \frac{- 1}{2 λ + 1} {\frac{(2 λ + 1) t^{2} + 2 t + 1 - 2 λ - 2 t - 1 + 2 λ - 2 λ - 1}{(2 λ + 1) t^{2} + 2 t + 1 - 2 λ}}$ $= - \frac{1}{2 λ + 1} {1 - \frac{2 t + 2}{(2 λ + 1) t^{2} + 2 t + 1 - 2 λ}}$ $= - \frac{1}{2 λ + 1} - \frac{2 t + 2}{(2 λ + 1) {(2 λ + 1) t^{2} + 2 t + 1 - 2 λ}}$ $. . . . b e c o n t i n u e d . . .$
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