let f(x) =∫_(−∞) ^(+∞) (dt/((t^2 +ixt −1))) with ∣x∣>2 (i^2 =−1) 1) extract Re(f(x)) and Im(f(x)) 2) calculate f(x) 3) find olso g(x) =∫_(−∞) ^(+∞) (t/((t^2 +ixt −1)^2 ))dt 4) find values of integrals ∫_(−∞) ^(+∞) (dt/((t^2 +3it −1))) and ∫_(−∞) ^(+∞) ((tdt)/((t^2 +3it −1)^2 )) 5) give f^((n)) (x) at form of integrals.


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Question Number 63508 by mathmax by abdo last updated on 05/Jul/19

$l e t f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{(t^{2} + i x t - 1)} w i t h ∣ x ∣> 2 (i^{2} = - 1)$ $1) e x t r a c t R e (f (x)) a n d I m (f (x))$ $2) c a l c u l a t e f (x)$ $3) f i n d o l s o g (x) = \int_{- \infty}^{+ \infty} \frac{t}{{(t^{2} + i x t - 1)}^{2}} d t$ $4) f i n d v a l u e s o f i n t e g r a l s \int_{- \infty}^{+ \infty} \frac{d t}{(t^{2} + 3 i t - 1)} a n d \int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} + 3 i t - 1)}^{2}}$ $5) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l s .$
Commented by mathmax by abdo last updated on 06/Jul/19

$1) w e h a v e f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{t^{2} - 1 + i x t} = \int_{- \infty}^{+ \infty} \frac{t^{2} - 1 - i x t}{{(t^{2} - 1)}^{2} + x^{2} t^{2}} d t$ $= \int_{- \infty}^{+ \infty} \frac{t^{2} - 1}{t^{4} - 2 t^{2} + 1 + x^{2} t^{2}} d t - i \int_{- \infty}^{+ \infty} \frac{x t}{t^{4} - 2 t^{2} + 1 + x^{2} t^{2}} d t$ $= \int_{- \infty}^{+ \infty} \frac{t^{2} - 1}{t^{4} + (x^{2} - 2) t^{2} + 1} - i \times 0 (t h e f u n c t i o n t \to \frac{t}{t^{4} + (x^{2} - 2) t^{2} + 1} i s o d d) \Rightarrow$ $R e (f (x)) = \int_{- \infty}^{+ \infty} \frac{t^{2} - 1}{t^{4} + (x^{2} - 2) t^{2} + 1} d t a n d I m (f (x)) = 0$
Commented by mathmax by abdo last updated on 06/Jul/19

$2) l e t w (z) = \frac{1}{z^{2} + i x z - 1} p o l e s o f w (z) ?$ $Δ = {(i x)}^{2} - 4 (- 1) = - x^{2} + 4 = 4 - x^{2} < 0 \Rightarrow Δ = {(i \sqrt{x^{2} - 4})}^{2} \Rightarrow$ $z_{1} = \frac{- i x + i \sqrt{x^{2} - 4}}{2} a n d z_{2} = \frac{- i x - i \sqrt{x^{2} - 4}}{2} \Rightarrow w (z) = \frac{1}{(z - z_{1}) (z - z_{2})}$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (w, z_{1})$ $R e s (w, z_{1}) = {l i m}_{z \to z_{1}} (z - z_{1}) w (z) = \frac{1}{z_{1} - z_{2}} = \frac{1}{i \sqrt{x^{2} - 4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π \frac{1}{i \sqrt{x^{2} - 4}} = \frac{2 π}{\sqrt{x^{2} - 4}} \Rightarrow f (x) = \frac{2 π}{\sqrt{x^{2} - 4}}$
Commented by mathmax by abdo last updated on 06/Jul/19

$3) b y d e r i v a t i o n w e h a v e f^{^{'}} (x) = - \int_{- \infty}^{+ \infty} \frac{i t}{{(t^{2} + i x t - 1)}^{2}} d t$ $= - i \int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} + i x t - 1)}^{2}} = - i g (x) \Rightarrow g (x) = i f^{^{'}} (x) w e h a v e$ $f (x) = \frac{2 π}{\sqrt{x^{2} - 4}} = 2 π {(x^{2} - 4)}^{- \frac{1}{2}} \Rightarrow f^{^{'}} (x) = 2 π (- \frac{1}{2}) (2 x) {(x^{2} - 4)}^{- \frac{3}{2}}$ $= \frac{- 2 π x}{(x^{2} - 4) \sqrt{x^{2} - 4}} \Rightarrow g (x) = \frac{- 2 π i x}{(x^{2} - 4) \sqrt{x^{2} - 4}} .$
Commented by mathmax by abdo last updated on 06/Jul/19

$4) w e h a v e p r o v e d t h a t \int_{- \infty}^{+ \infty} \frac{d t}{t^{2} + i x t - 1} = \frac{2 π}{\sqrt{x^{2} - 4}} i f ∣ x ∣> 2 \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{d t}{t^{2} + 3 i t - 1} = \frac{2 π}{\sqrt{3^{2} - 4}} = \frac{2 π}{\sqrt{5}} a l s o w e h a v e$ $\int_{- \infty}^{+ \infty} \frac{t}{{(t^{2} + i x t - 1)}^{2}} d t = \frac{- 2 π i x}{(x^{2} - 4) \sqrt{x^{2} - 4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} + 3 i t - 1)}^{2}} = \frac{- 6 i π}{5 \sqrt{5}} .$
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