let-f-sin-x-x-2-2-x-1-2-dx-with-lt-1-1-find-the-value-of-f-2-calculate-sin-x-2-x-2-x-1-2-dx-3-find-A-

Question Number 52683 by maxmathsup by imad last updated on 11/Jan/19

l e t f (λ) = \int_{- \infty}^{+ \infty} \frac{s i n (λ x)}{{(x^{2} + 2 λ x + 1)}^{2}} d x w i t h ∣ λ ∣< 1

1) f i n d t h e v a l u e o f f (λ)

2) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{s i n (\frac{x}{2})}{{(x^{2} + x + 1)}^{2}} d x

3) f i n d A (θ) = \int_{- \infty}^{+ \infty} \frac{s i n ((c o s θ) x)}{{(x^{2} + 2 c o s θ x + 1)}^{2}} t h a t w e s u p p o s e 0 < θ < \frac{π}{2}

Commented by maxmathsup by imad last updated on 13/Jan/19

1) f (λ) = I m (\int_{- \infty}^{+ \infty} \frac{e^{i λ x}}{{(x^{2} + 2 λ x + 1)}^{2}} d x) l e t c o n s d e r t h e c o m p l e x f u n c t i o n

φ (z) = \frac{e^{i λ z}}{{(x^{2} + 2 λ x + 1)}^{2}} p o l e s o f φ ? l e t d e t e r m i n e r o o t s o f x^{2} + 2 λ x + 1

Δ^{^{'}} = λ^{2} - 1 < 0 \Rightarrow Δ^{^{'}} = - (1 - λ^{2}) = {(i \sqrt{1 - λ^{2}})}^{2} \Rightarrow z_{1} = - λ + i \sqrt{1 - λ^{2}}

a n d z_{2} = - λ - i \sqrt{1 - λ^{2}} \Rightarrow φ h a v e d o u b l e p o l e s z_{1} a n d z_{2}

φ (z) = \frac{e^{i λ z}}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} r e s i d u s t h e o r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1}) b u t

R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}

= {l i m}_{z \to z_{1}} {\frac{e^{i λ z}}{{(z - z_{2})}^{2}}}^{(1)}

= {l i m}_{z \to z_{1}} \frac{i λ e^{i λ z} {(z - z_{2})}^{2} - 2 (z - z_{2}) e^{i λ z}}{{(z - z_{2})}^{4}} = {l i m}_{z \to z_{1}} \frac{(i λ (z - z_{2}) - 2) e^{i λ z}}{{(z - z_{2})}^{3}}

= \frac{(i λ (z_{1} - z_{2}) - 2) e^{{i z}_{1}}}{{(z_{1} - z_{2})}^{3}} = \frac{(i λ (2 i \sqrt{1 - λ^{2}}) - 2) e^{i λ z_{1}}}{{(2 i \sqrt{1 - λ^{2}})}^{3}}

= \frac{(- 2 λ \sqrt{1 - λ^{2}} - 2) e^{i λ (- λ + i \sqrt{1 - λ^{2}})}}{- 8 i (1 - λ^{2}) \sqrt{1 - λ^{2}}} = \frac{(1 + λ \sqrt{1 - λ^{2}}) e^{- i λ^{2}} e^{- λ \sqrt{1 - λ^{2}}}}{4 i (1 - λ^{2}) \sqrt{1 - λ^{2}}}

\Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(1 + λ \sqrt{1 - λ^{2}}) e^{- λ \sqrt{1 - λ^{2}}}}{4 i (1 - λ^{2}) \sqrt{1 - λ^{2}}} e^{- i λ^{2}}

= \frac{π}{2} \frac{(1 + λ \sqrt{1 - λ^{2}}) e^{- λ \sqrt{1 - λ^{2}}}}{(1 - λ^{2}) \sqrt{1 - λ^{2}}} (c o s (λ^{2}) - i s i n (λ^{2})) \Rightarrow

f (λ) = - \frac{π}{2} s i n (λ^{2}) \frac{(1 + λ \sqrt{1 - λ^{2}}) e^{- λ \sqrt{1 - λ^{2}}}}{(1 - λ^{2}) \sqrt{1 - λ^{2}}} .

Commented by maxmathsup by imad last updated on 13/Jan/19

2) \int_{- \infty}^{+ \infty} \frac{s i n (\frac{x}{2})}{{(x^{2} + x + 1)}^{2}} d x = f (\frac{1}{2}) = - \frac{π}{2} s i n (\frac{1}{4}) \frac{(1 + \frac{1}{2} \sqrt{\frac{3}{4}}) e^{- \frac{1}{2} \sqrt{\frac{3}{4}}}}{\frac{3}{4} \sqrt{\frac{3}{4}}}

= - \frac{π}{2} s i n (\frac{1}{4}) \frac{(1 + \frac{\sqrt{3}}{4}) e^{- \frac{\sqrt{3}}{4}}}{\frac{3 \sqrt{3}}{8}} = - 4 π s i n (\frac{1}{4}) \frac{(1 + \frac{\sqrt{3}}{4}) e^{- \frac{\sqrt{3}}{4}}}{3 \sqrt{3}} .

Commented by Abdo msup. last updated on 13/Jan/19

3) A (θ) = f (c o s θ) = - \frac{π}{2} s i n ({c o s}^{2} θ) \frac{(1 + c o s θ s i n θ) e^{- c o s θ s i n θ}}{{s i n}^{3} θ}