let-A-0-dx-x-2-i-2-i-2-1-1-calculate-A-2-let-R-Re-A-and-I-Im-A-find-the-value-of-R-and-I-

Question Number 62141 by maxmathsup by imad last updated on 15/Jun/19

l e t A = \int_{0}^{+ \infty} \frac{d x}{{(x^{2} - i)}^{2}} (i^{2} = - 1)

1) c a l c u l a t e A

2) l e t R = R e (A) a n d I = I m (A)

f i n d t h e v a l u e o f R a n d I .

Commented by maxmathsup by imad last updated on 16/Jun/19

1) w e h a v e 2 A = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - i)}^{2}} l e t w (z) = \frac{1}{{(z^{2} - i)}^{2}} \Rightarrow

w (z) = \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 w a r e \bar{+} e^{\frac{i π}{4}} (d o u b l e s)

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, e^{\frac{i π}{4}})

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

= - \frac{1}{4} {- c o s (\frac{π}{4}) - i s i n (\frac{π}{4})} = \frac{1}{4} \frac{\sqrt{2}}{2} + i \frac{1}{4} \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{8} + i \frac{\sqrt{2}}{8} \Rightarrow

\int_{- \infty}^{+ \infty} w (z) d z = 2 i π {\frac{\sqrt{2}}{8} + i \frac{\sqrt{2}}{8}} = \frac{i π \sqrt{2}}{4} - \frac{π \sqrt{2}}{4} = 2 A \Rightarrow

A = - \frac{π \sqrt{2}}{8} + \frac{i π \sqrt{2}}{8}

2) w e h a v e A = \int_{0}^{\infty} \frac{d x}{{(x^{2} - i)}^{2}} = \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 + 2 {i x}^{2}}{x^{8} + 2 x^{4} + 1} d x

= \int_{0}^{\infty} \frac{x^{4} - 1}{x^{8} + 2 x^{4} + 1} d x + i \int_{0}^{\infty} \frac{2 x^{2}}{x^{8} + 2 x^{4} + 1} d x \Rightarrow

R = \int_{0}^{\infty} \frac{x^{4} - 1}{x^{8} + 2 x^{4} + 1} a n d I = \int_{0}^{\infty} \frac{2 x^{2}}{x^{8} + 2 x^{4} + 1} d x \Rightarrow

R = - \frac{π \sqrt{2}}{8} a n d I = \frac{π \sqrt{2}}{8} .