let-give-I-0-dx-x-2-i-2-1-extract-Re-I-and-Im-I-2-find-the-value-of-I-3-calculate-Re-I-and-Im-I-

Question Number 36167 by abdo mathsup 649 cc last updated on 29/May/18

l e t g i v e I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + i)}^{2}}

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

2) f i n d t h e v a l u e o f I

3) c a l c u l a t e R e (I) a n d I m (I) .

Commented by maxmathsup by imad last updated on 22/Aug/18

1) w e h a v e I = \int_{0}^{\infty} \frac{{(x^{2} - i)}^{2}}{{(x^{2} + i)}^{2} {(x^{2} - i)}^{2}} d x

= \int_{0}^{\infty} \frac{x^{4} - 2 {i x}^{2} - 1}{{(x^{4} + 1)}^{2}} d x = \int_{0}^{\infty} \frac{x^{4} - 1}{{(x^{4} + 1)}^{2}} d x + i \int_{0}^{\infty} \frac{- 2 x^{2}}{{(x^{4} + 1)}^{2}} d x \Rightarrow

R e (I) = \int_{0}^{\infty} \frac{x^{4} - 1}{{(x^{4} + 1)}^{2}} d x a n d I m (I) = \int_{0}^{\infty} \frac{- 2 x^{2}}{{(x^{4} + 1)}^{2}} d x

Commented by maxmathsup by imad last updated on 22/Aug/18

2) w e h a v e 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + i)}^{2}} l e t c o n s i d e r t h e c o m p l e x f u n c t i o n

φ (z) = \frac{1}{{(z^{2} + i)}^{2}} \Rightarrow φ (z) = \frac{1}{{(z^{2} - {(\sqrt{- i})}^{2})}^{2}} = \frac{1}{{(z - \sqrt{- i})}^{2} {(z + \sqrt{- i})}^{2}}

= \frac{1}{{(z - e^{- \frac{i π}{4}})}^{2} {(z + e^{- \frac{i π}{4}})}^{2}} s o t h e p o l e s o f φ a r e \bar{+} e^{- \frac{i π}{4}} (d o u b l e s) \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, - e^{- \frac{i π}{4}}) b u t

R e s (φ, e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z + e^{- \frac{i π}{4}})}^{2} φ (z)}^{(1)}

= {l i m}_{z \to - e^{- \frac{i π}{4}}} {{(z - e^{- \frac{i π}{4}})}^{- 2}}^{(1)} = {l i m}_{z \to - e^{- \frac{i π}{4}}} - 2 {(z - e^{- \frac{i π}{4}})}^{- 3}

= - 2 {(- 2 e^{- \frac{i π}{4}})}^{- 3} = \frac{- 2}{{(- 2)}^{3}} e^{i \frac{3 π}{4}} = \frac{1}{4} e^{\frac{i 3 π}{4}} = \frac{1}{4} {c o s (\frac{3 π}{4}) + i s i n (\frac{3 π}{4})}

= \frac{1}{4} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{4} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = \frac{i π}{2} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = 2 I \Rightarrow

I = \frac{i π}{4} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = - \frac{π}{4 \sqrt{2}} - \frac{π}{4 \sqrt{2}} i

3) w e h a v e I = R e (I) + i I m (I) \Rightarrow R e (I) = - \frac{π}{4 \sqrt{2}} a n d I m (I) = - \frac{π}{4 \sqrt{2}}