let A(t) = ∫_(−∞) ^(+∞) ((sin(xt))/(( x +1+i)^2 )) dx with t from R 2) calculate A(t) 2) extract Re(A(t)) and Im(A(t)) 3) find the value of ∫


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Question Number 36168 by abdo mathsup 649 cc last updated on 29/May/18

$l e t A (t) = \int_{- \infty}^{+ \infty} \frac{s i n (x t)}{{(x + 1 + i)}^{2}} d x w i t h t f r o m R$ $2) c a l c u l a t e A (t)$ $2) e x t r a c t R e (A (t)) a n d I m (A (t))$ $3) f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{c o s (3 x)}{{(x + 1 + i)}^{2}} d x$
Commented by maxmathsup by imad last updated on 20/Aug/18
$2) we have A(t) =∫_(−∞) ^(+∞) ((sin(xt)(x+1−i)^2 )/((x+1+i)^2 (x+1−i)^2 ))dx = ∫_(−∞) ^(+∞) ((sin(xt){x^2 +2x(1−i) +(1−i)^2 })/({(x+1)^2 +1}^2 )) dx =∫_(−∞) ^(+∞) ((sin(xt){ x^2 +2x−2xi −2i })/({(x+1)^z +1}^2 ))dx =∫_(−∞) ^(+∞) (((x^2 +2x)sin(xt))/({(x+1)^2 +1}^2 ))dx + i ∫_(−∞) ^(+∞) (((−2x−2)sin(xt))/({(x+1)^2 +1}^2 ))dx ⇒ Re(A(x)) =∫_(−∞) ^(+∞) (((x^2 +2x)sin(xt))/({(x+1)^2 +1}^2 ))dx and Im (A(x)) = ∫_(−∞) ^(+∞) (((−2x−2)sin(xt))/({(x+1)^2 +1}^2 )) dx$
$2) w e h a v e A (t) = \int_{- \infty}^{+ \infty} \frac{s i n (x t) {(x + 1 - i)}^{2}}{{(x + 1 + i)}^{2} {(x + 1 - i)}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{s i n (x t) {x^{2} + 2 x (1 - i) + {(1 - i)}^{2}}}{{{(x + 1)}^{2} + 1}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{s i n (x t) {x^{2} + 2 x - 2 x i - 2 i}}{{{(x + 1)}^{z} + 1}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{(x^{2} + 2 x) s i n (x t)}{{{(x + 1)}^{2} + 1}^{2}} d x + i \int_{- \infty}^{+ \infty} \frac{(- 2 x - 2) s i n (x t)}{{{(x + 1)}^{2} + 1}^{2}} d x \Rightarrow$ $R e (A (x)) = \int_{- \infty}^{+ \infty} \frac{(x^{2} + 2 x) s i n (x t)}{{{(x + 1)}^{2} + 1}^{2}} d x a n d$ $I m (A (x)) = \int_{- \infty}^{+ \infty} \frac{(- 2 x - 2) s i n (x t)}{{{(x + 1)}^{2} + 1}^{2}} d x$
Commented by maxmathsup by imad last updated on 21/Aug/18
$3) let I = ∫_(−∞) ^(+∞) ((cos(3x))/((x+1+i)^2 ))dx ⇒I = ∫_(−∞) ^(+∞) ((cos(3x)(x+1−i)^2 )/((x+1+i)^2 (x+1−i)^2 ))dx = ∫_(−∞) ^(+∞) ((cos(3x){x^2 +2x(1−i) +(1−i)^2 ))/({(x+1)^2 +1}^2 )) dx = ∫_(−∞) ^(+∞) ((cos(3x){x^2 +2x −2xi −2i})/({(x+1)^2 +1}^2 )) dx =∫_(−∞) ^(+∞) (((x^2 +2x)cos(3x))/({(x+1)^2 +1}^2 )) dx + i∫_(−∞) ^(+∞) (((−2x−2)cos(3x))/({(x+1)^2 +1}^2 )) dx =H +iK let find H H = Re( ∫_(−∞) ^(+∞) (((x^2 +2x)e^(i3x) )/({(x+1)^2 +1}^2 ))dx) cha7gement x+1 =t give ∫_(−∞) ^(+∞) (((x^2 +2x)e^(i3x) )/({(x+1)^2 +1}^2 ))dx = ∫_(−∞) ^(+∞) (({(t−1)^2 +2(t−1)}e^(i3(t−1)) )/((t^2 +1)^2 )) = ∫_(−∞) ^(+∞) (({t^2 −2t +1 +2t−2} e^(i3(t−1)) )/((t^2 +1)^2 ))dt=e^(−3i) ∫_(−∞) ^(+∞) (((t^2 −1)e^(i3t) )/((t^2 +1)^2 ))dt let ϕ(z) =(((z^2 −1)e^(i3z) )/((z^2 +1)^2 )) we have ϕ(z) =(((z^2 −1) e^(3iz) )/((z−i)^2 (z+i)^2 )) the poles of ϕ are i and −i(doubles) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) { (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(((z^2 −1)e^(3iz) )/((z+i)^2 ))}^((1)) = lim_(z→i) (((2z e^(3iz) +3i(z^2 −1)e^(3iz) )(z+i)^2 −2(z+i)(z^2 −i)e^(3iz) )/((z+i)^4 )) =lim_(z→i) (((2z +3iz^2 −3i)e^(3iz) (z+i) −2(z^2 −i) e^(3iz) )/((z+i)^3 )) =(((2i−3i −3i)e^(−3) (2i) −2(−1−i)e^(−3) )/((2i)^3 )) = ((−4i(2i)e^(−3) +2(1+i)e^(−3) )/(−8i)) =−(((10 +2i)e^(−3) )/(8i)) =((i(5 +i)e^(−3) )/4) ⇒ e^(−3i) ∫_(−∞) ^(+∞) ϕ(z)dz =e^(−3i ) {2iπ ((i(5+i)e^(−3) )/4)} =e^(−3i) (−(π/2)(5+i)) =−(π/2)(5+i)( cos(3)−isin(3)) =−(π/2){5 cos(3)−5i sin(3) +i cos(3) +sin(3)} H =Re( ∫...) =−(π/2)( 5 cos(3) +sin(3)) let find k we have −K =Re( ∫_(−∞) ^(+∞) (((2x+2) e^(i3x) )/({(x+1)^2 +1}^2 ))dx) changement x+1 =t give ∫_(−∞) ^(+∞) (((2x+2) e^(i3x) )/({(x+1)^2 +1}^2 ))dx =∫_(−∞) ^(+∞) (({2(t−1) +2}e^(i3(t−1)) )/({t^2 +1}^2 ))dt = ∫_(−∞) ^(+∞) ((2t e^(−3i) e^(i3t) )/((t^2 +1)^2 ))dt = 2 e^(−3i) ∫_(−∞) ^(+∞) ((t e^(3it) )/((t^2 +1)^2 ))dt let ϕ(z) =((z e^(3iz) )/((z^2 +1)^2 )) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Re(ϕ,i)....be continued...$
$3) l e t I = \int_{- \infty}^{+ \infty} \frac{c o s (3 x)}{{(x + 1 + i)}^{2}} d x \Rightarrow I = \int_{- \infty}^{+ \infty} \frac{c o s (3 x) {(x + 1 - i)}^{2}}{{(x + 1 + i)}^{2} {(x + 1 - i)}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{c o s (3 x) {x^{2} + 2 x (1 - i) + {(1 - i)}^{2})}{{{(x + 1)}^{2} + 1}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{c o s (3 x) {x^{2} + 2 x - 2 x i - 2 i}}{{{(x + 1)}^{2} + 1}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{(x^{2} + 2 x) c o s (3 x)}{{{(x + 1)}^{2} + 1}^{2}} d x + i \int_{- \infty}^{+ \infty} \frac{(- 2 x - 2) c o s (3 x)}{{{(x + 1)}^{2} + 1}^{2}} d x = H + i K$ $l e t f i n d H$ $H = R e (\int_{- \infty}^{+ \infty} \frac{(x^{2} + 2 x) e^{i 3 x}}{{{(x + 1)}^{2} + 1}^{2}} d x) c h a 7 g e m e n t x + 1 = t g i v e$ $\int_{- \infty}^{+ \infty} \frac{(x^{2} + 2 x) e^{i 3 x}}{{{(x + 1)}^{2} + 1}^{2}} d x = \int_{- \infty}^{+ \infty} \frac{{{(t - 1)}^{2} + 2 (t - 1)} e^{i 3 (t - 1)}}{{(t^{2} + 1)}^{2}}$ $= \int_{- \infty}^{+ \infty} \frac{{t^{2} - 2 t + 1 + 2 t - 2} e^{i 3 (t - 1)}}{{(t^{2} + 1)}^{2}} d t = e^{- 3 i} \int_{- \infty}^{+ \infty} \frac{(t^{2} - 1) e^{i 3 t}}{{(t^{2} + 1)}^{2}} d t$ $l e t φ (z) = \frac{(z^{2} - 1) e^{i 3 z}}{{(z^{2} + 1)}^{2}} w e h a v e φ (z) = \frac{(z^{2} - 1) e^{3 i z}}{{(z - i)}^{2} {(z + i)}^{2}}$ $t h e p o l e s o f φ a r e i a n d - i (d o u b l e s) \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t$ $R e s (φ, i) = {l i m}_{z \to i} {{(z - i)}^{2} φ (z)}^{(1)} = {l i m}_{z \to i} {\frac{(z^{2} - 1) e^{3 i z}}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{(2 z e^{3 i z} + 3 i (z^{2} - 1) e^{3 i z}) {(z + i)}^{2} - 2 (z + i) (z^{2} - i) e^{3 i z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(2 z + 3 {i z}^{2} - 3 i) e^{3 i z} (z + i) - 2 (z^{2} - i) e^{3 i z}}{{(z + i)}^{3}}$ $= \frac{(2 i - 3 i - 3 i) e^{- 3} (2 i) - 2 (- 1 - i) e^{- 3}}{{(2 i)}^{3}} = \frac{- 4 i (2 i) e^{- 3} + 2 (1 + i) e^{- 3}}{- 8 i}$ $= - \frac{(10 + 2 i) e^{- 3}}{8 i} = \frac{i (5 + i) e^{- 3}}{4} \Rightarrow$ $e^{- 3 i} \int_{- \infty}^{+ \infty} φ (z) d z = e^{- 3 i} {2 i π \frac{i (5 + i) e^{- 3}}{4}} = e^{- 3 i} (- \frac{π}{2} (5 + i))$ $= - \frac{π}{2} (5 + i) (c o s (3) - i s i n (3))$ $= - \frac{π}{2} {5 c o s (3) - 5 i s i n (3) + i c o s (3) + s i n (3)}$ $H = R e (\int . . .) = - \frac{π}{2} (5 c o s (3) + s i n (3)) l e t f i n d k w e h a v e$ $- K = R e (\int_{- \infty}^{+ \infty} \frac{(2 x + 2) e^{i 3 x}}{{{(x + 1)}^{2} + 1}^{2}} d x) c h a n g e m e n t x + 1 = t g i v e$ $\int_{- \infty}^{+ \infty} \frac{(2 x + 2) e^{i 3 x}}{{{(x + 1)}^{2} + 1}^{2}} d x = \int_{- \infty}^{+ \infty} \frac{{2 (t - 1) + 2} e^{i 3 (t - 1)}}{{t^{2} + 1}^{2}} d t$ $= \int_{- \infty}^{+ \infty} \frac{2 t e^{- 3 i} e^{i 3 t}}{{(t^{2} + 1)}^{2}} d t = 2 e^{- 3 i} \int_{- \infty}^{+ \infty} \frac{t e^{3 i t}}{{(t^{2} + 1)}^{2}} d t l e t$ $φ (z) = \frac{z e^{3 i z}}{{(z^{2} + 1)}^{2}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e (φ, i) . . . . b e c o n t i n u e d . . .$
Commented by math khazana by abdo last updated on 21/Aug/18

$H = - \frac{π}{2} e^{- 3} (5 c o s (3) + s i n (3)) .$
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