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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) ★ 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}} .