let-I-cos-x-1-ix-2-dx-1-extract-Re-I-and-Im-I-2-calculate-I-3-conclude-the-values-of-Re-I-and-Im-I-

Question Number 39370 by maxmathsup by imad last updated on 05/Jul/18

l e t I (λ) = \int_{- \infty}^{+ \infty} \frac{c o s (λ x)}{{(1 + i x)}^{2}} d x

1) e x t r a c t R e (I (λ)) a n d I m (I (λ))

2) c a l c u l a t e I (λ)

3) c o n c l u d e t h e v a l u e s o f R e (I (λ)) a n d I m (I (λ)) .

Commented by abdo mathsup 649 cc last updated on 08/Jul/18

1) w e h a v e I (λ) = \int_{- \infty}^{+ \infty} \frac{c o s (λ x) {(1 - i x)}^{2}}{{(1 + i x)}^{2} {(1 - i x)}^{2}}

= \int_{- \infty}^{+ \infty} \frac{c o s (λ x) {(1 - i x)}^{2}}{{(1 + x^{2})}^{2}} d x

= \int_{- \infty}^{+ \infty} \frac{c o s (λ x) (1 - 2 i x - x^{2})}{{(1 + x^{2})}^{2}} d x

= \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x) - 2 i x c o s (λ x)}{{(1 + x^{2})}^{2}} d x

= \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x)}{{(1 + x^{2})}^{2}} d x - i \int_{- \infty}^{+ \infty} \frac{x c o s (λ x)}{{(1 + x^{2})}^{2}} d x \Rightarrow

R e (I (λ)) = \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x)}{{(1 + x^{2})}^{2}} d x a n d

I m (I (λ)) = - \int_{- \infty}^{+ \infty} \frac{x c o s (λ x)}{{(1 + x^{2})}^{2}} d x = 0

2) I (λ) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i λ x}}{{(1 + i x)}^{2}} d x) l e t

φ (z) = \frac{e^{i λ z}}{{(1 + i z)}^{2}} p o l e s o f φ ?

φ (z) = \frac{e^{i λ z}}{{(i z - i^{2})}^{2}} = \frac{e^{i λ z}}{(- 1) {(z - i)}^{2}} = - \frac{e^{i λ z}}{{(z - i)}^{2}}

s o i s a d o u b l e p o l e f o r φ

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)

R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}

= {l i m}_{z \to i} {- e^{i λ z}}^{(1)} = {l i m}_{z \to i} - i λ e^{i λ z}

= - i λ e^{- λ} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (- i λ e^{- λ}) = 2 π λ e^{- λ} \Rightarrow

I (I (λ)) = 2 π λ e^{- λ}

3) R e (I (λ)) = \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x)}{{(1 + x^{2})}^{2}} d x = 2 π λ e^{- λ}

a n d I m (I (λ)) = 0