let-I-n-0-pi-vos-nt-1-2-cost-2-dt-1-calculate-I-0-and-I-1-2-find-relation-between-I-n-1-I-n-and-I-n-1-3-calculate-I-n-

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) = \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}}

\dots . b e c o n t i n u e d \dots