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Question Number 25851 by abdo imad last updated on 15/Dec/17

f i n d t h e v a l u e o f i n t e g r a l \int_{R} {(z - a)}^{- 1} d z w i t h a f r o m C a p l l y

t h i s r e s u l t t o f i n d t h e v a l u e o f \int_{0}^{\infty} {(2 + x^{4_{}})}^{- 1} d x .

Commented by abdo imad last updated on 21/Dec/17

l e t p u t I (ξ) = \int_{- ξ}^{ξ} \frac{d x}{x - a} a n d a = α + i β a n d β n o t 0

I (ξ) = \int_{- ξ}^{ξ} \frac{x - α + i β}{{(x - α)}^{2} + β^{2^{}}} d x = A (ξ) + B (ξ) w h e r e

A (ξ) = \int_{- ξ}^{ξ} \frac{x - α}{{(x - α)}^{2} + β^{2}} d x a n d B (ξ) = i β \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}}

b u t A (ξ) = \frac{1}{2} l n (\frac{{(ξ - α)}^{2} + β^{2}}{{(ξ + α)}^{2} + β^{2}}) a n d l i m_{ξ - >\propto} A (ξ) = 0

a n d b y t h e c h a n g e m e n t x - α = β t

B (ξ) = i β \int_{\frac{- ξ - α}{β}}^{\frac{ξ - α}{β}} \frac{β}{β^{2} t^{2} + β^{2}} d t = i (a r t a n (\frac{ξ - α}{β}) + a r t a n (\frac{ξ + α}{β}))

- - > i f β > 0 l i m B {(ξ)}_{ξ - >\propto} = i π

i f β < 0 l i m B {(ξ)}_{ξ - >\propto} = - i π \dots l o o k t h a t β = i m (a)

v a l u e o f I = \int_{0}^{\infty} \frac{d x}{2 + x^{4}} I = \frac{1}{2} \int_{R} \frac{d x}{2 + x^{4}} a n d b y c h a n g e m e n t

x = 2^{\frac{1}{4}} t \dots I = \frac{2^{\frac{1}{4}}}{4} \int_{R} \frac{d t}{t^{4} + 1} l e t f i n d t h e p o l e s o f f (z) = \frac{1}{z^{4} + 1}

t h e r o o t s o f z^{4} + 1 = 0 a r e t h e c o m p l e x z_{k} = e^{i (2 k + 1) \frac{π}{4}} k f r o m [[0.3]]

z_{0} = e^{i \frac{π}{4}} . . z_{1} = e^{i \frac{3 π}{4}} . . z_{2} = e^{i \frac{5 π}{4}} . . z_{3} = e^{i \frac{7 π_{}}{4}}

f (z) = \sum_{k = 0}^{k = 3} \frac{λ_{k}}{z - z_{k}} a n d λ_{k} = \frac{1}{d e r (1 + z^{4})} = - \frac{1}{4} z_{k} (d e r m e a n s d e r i v a t i v e)

\int_{R} f (z) d z = \frac{- 1}{4} (\int_{R} \frac{z_{0}}{z - z_{0}} d z + \int_{R} \frac{z_{1}}{z - z_{1}} d z + \int_{R} \frac{z_{2}}{z - z_{2}} d z + \int_{R} \frac{z_{3}}{z - z_{3}} d z)

= \frac{- 1}{4} (i π z_{0} + i π z_{1} - i π z_{2} - i π z_{3}) b e c a u s e i m (z_{0}) > 0

i m (z_{1}) > 0 . . i m (z_{2}) < 0 i m (z_{3}) < 0

\int_{R} f (z) d z = \frac{- i π}{4} (z_{0} + z_{1} + z_{2} + z_{3}) = \frac{- i π}{4} (z_{0} - z_{0}^{-} + z_{0} - z_{0}^{-})

= \frac{- i π}{2} (2 i) i m (z_{0}) = \frac{π \sqrt{2}}{2} - - - > \int_{0}^{\infty} \frac{d x}{2 + x^{4}} = 2^{\frac{1}{4}} . 4^{- 1} . \frac{π \sqrt{2}}{2}

- - - > \int_{R} \frac{d x}{2 + x^{4}} = \frac{π \sqrt{2} . 2^{\frac{1}{4}}}{8} \dots