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-cos-3x-x-1-i-2-dx-

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) 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) 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 \dots) = - \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) \dots . b e c o n t i n u e d \dots

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)) .