calculate f(α) =∫_(−∞) ^(+∞) (dx/(x^2 +2x cosα +1)) 2) calculate g(α)=∫_(−∞) ^(+∞) ((sinα)/((x^2 +2x cosα+1)^2 ))dx 3) find f^((n)) (α) with n integr natural . 4) calculate ∫_(−∞) ^(+∞) (dx/(x^2 +x +1)) and ∫


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 47295 by maxmathsup by imad last updated on 07/Nov/18

$c a l c u l a t e f (α) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 x c o s α + 1}$ $2) c a l c u l a t e g (α) = \int_{- \infty}^{+ \infty} \frac{s i n α}{{(x^{2} + 2 x c o s α + 1)}^{2}} d x$ $3) f i n d f^{(n)} (α) w i t h n i n t e g r n a t u r a l .$ $4) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + x + 1} a n d \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{2}}$
Commented by maxmathsup by imad last updated on 08/Nov/18

$s o r r y t h e Q i s c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{2 x s i n α}{{(x^{2} + 2 x c o s α + 1)}^{2}} d x .$
Commented by maxmathsup by imad last updated on 08/Nov/18

$l e t r e c t f y g (\frac{π}{3}) = \int_{- \infty}^{+ \infty} \frac{2 x s i n (\frac{π}{3})}{{(x^{2} + x + 1)}^{2}} d x = \int_{- \infty}^{+ \infty} \frac{x \sqrt{3}}{{(x^{2} + x + 1)}^{2}} d x \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{x}{{(x^{2} + x + 1)}^{2}} d x = \frac{1}{\sqrt{3}} g (\frac{π}{3}) = - \frac{1}{\sqrt{3}} \frac{π c o s (\frac{π}{3})}{{s i n}^{2} (\frac{π}{3})} = \frac{- π}{2 \sqrt{3}} \frac{4}{3} = \frac{- 2 π}{3 \sqrt{3}} .$
Commented by maxmathsup by imad last updated on 08/Nov/18

$2) w e h a v e f^{^{'}} (α) = \int_{- \infty}^{+ \infty} \frac{\partial}{\partial α} (\frac{1}{x^{2} + 2 x c o s α + 1}) d x$ $= \int_{- \infty}^{+ \infty} \frac{2 x s i n α}{{(x^{2} + 2 x c o s α + 1)}^{2}} d x = g (α) \Rightarrow g (α) = f^{^{'}} (α)$ $c a s e 1 s i n α > 0 \Rightarrow f (α) = \frac{π}{s i n α} \Rightarrow f^{^{'}} (α) = - \frac{π c o s α}{{s i n}^{2} α} = g (α)$ $c a s e z s i n α < 0 \Rightarrow f (α) = - \frac{π}{s i n α} \Rightarrow f^{^{'}} (α) = \frac{π c o s α}{{s i n}^{2} α} .$
Commented by maxmathsup by imad last updated on 08/Nov/18

$1) l e t d e t e r m i n e t h e p o l e s o f φ (z) = \frac{1}{z^{2} + 2 z c o s α + 1}$ $Δ^{^{'}} = {c o s}^{2} - 1 = - {s i n}^{2} α = {(i s i n α)}^{2} \Rightarrow z_{1} = - c o s α + i s i n α = - e^{- i α}$ $z_{2} = - c o s α - i s i n α = - e^{i α} \Rightarrow φ (z) = \frac{1}{(z + e^{i α}) (z + e^{- i α})} = \frac{1}{(z - z_{1}) (z - z_{2})}$ $c a s e 1 s i n α > 0 r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})$ $= 2 i π . \frac{1}{z_{1} - z_{2}} = 2 i π \frac{1}{2 i s i n α} = \frac{π}{s i n α} \Rightarrow f (α) = \frac{π}{s i n α}$ $c a s e 2 s i n α < 0 \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{2}) = 2 i π \frac{1}{z_{2} - z_{1}} = 2 i π \frac{1}{- 2 i s i n α}$ $= - \frac{π}{s i n α} \Rightarrow f (α) = - \frac{π}{s i n α} .$
Commented by maxmathsup by imad last updated on 08/Nov/18

$4) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x d x}{{(x^{2} + x + 1)}^{2}} .$
Commented by maxmathsup by imad last updated on 08/Nov/18
$4) ★we have x^2 +x +1 =x^2 +2(1/2)x +1 =x^2 +2x cos((π/3))+1 ⇒ ∫_(−∞) ^(+∞) (dx/(x^2 +x+1)) =f((π/3)) =(π/(sin((π/3)))) =((2π)/(√3)) another way ∫_(−∞) ^(+∞) (dx/(x^2 +x+1)) =∫_(−∞) ^(+∞) (dx/((x+(1/2))^2 +(3/4))) =_(x+(1/2)=((√3)/2)t) ∫_(−∞) ^(+∞) (1/((3/(4 ))(1+t^2 ))) ((√3)/2)dt =(4/3) ((√3)/2) ∫_(−∞) ^(+∞) (dt/(1+t^2 )) =(2/(√3)) [arctan(t)]_(−∞) ^(+∞) =(2/(√3)){(π/2)+(π/2)} =((2π)/(√3)) . ★ we have g((π/3))=∫_(−∞) ^(+∞) ((sin((π/3)))/((x^2 +x+1)^2 )) dx ⇒ ∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^2 )) =(1/(sin((π/3)))) g((π/3)) =(2/(√3)) ((π cos((π/3)))/(sin^2 ((π/3)))) =(π/(√3)) (1/(3/4)) =((4π)/(3(√3))) .$
$4) ★ w e h a v e x^{2} + x + 1 = x^{2} + 2 \frac{1}{2} x + 1 = x^{2} + 2 x c o s (\frac{π}{3}) + 1 \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + x + 1} = f (\frac{π}{3}) = \frac{π}{s i n (\frac{π}{3})} = \frac{2 π}{\sqrt{3}}$ $a n o t h e r w a y \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + x + 1} = \int_{- \infty}^{+ \infty} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$ $=_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int_{- \infty}^{+ \infty} \frac{1}{\frac{3}{4} (1 + t^{2})} \frac{\sqrt{3}}{2} d t = \frac{4}{3} \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} \frac{d t}{1 + t^{2}} = \frac{2}{\sqrt{3}} {[a r c t a n (t)]}_{- \infty}^{+ \infty}$ $= \frac{2}{\sqrt{3}} {\frac{π}{2} + \frac{π}{2}} = \frac{2 π}{\sqrt{3}} .$ $★ w e h a v e g (\frac{π}{3}) = \int_{- \infty}^{+ \infty} \frac{s i n (\frac{π}{3})}{{(x^{2} + x + 1)}^{2}} d x \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{2}} = \frac{1}{s i n (\frac{π}{3})} g (\frac{π}{3}) = \frac{2}{\sqrt{3}} \frac{π c o s (\frac{π}{3})}{{s i n}^{2} (\frac{π}{3})} = \frac{π}{\sqrt{3}} \frac{1}{\frac{3}{4}} = \frac{4 π}{3 \sqrt{3}} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum