1-calculate-dx-x-a-with-a-C-2-find-the-values-of-0-dx-x-4-1-and-0-dx-x-6-1-by-using-the-decomposition-inside-C-x-

Question Number 65293 by mathmax by abdo last updated on 27/Jul/19

1) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{x - a} w i t h a \in C

2) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{d x}{x^{4} + 1} a n d \int_{0}^{\infty} \frac{d x}{x^{6} + 1}

b y u s i n g t h e d e c o m p o s i t i o n i n s i d e C (x) .

Commented by mathmax by abdo last updated on 01/Aug/19

1) l e t I (ξ) = \int_{- ξ}^{ξ} \frac{d x}{x - a} l e t a = α + i β w e h a v e \int_{- \infty}^{+ \infty} \frac{d x}{x - a} = {l i m}_{ξ \to + \infty} I (ξ)

I (ξ) = \int_{- ξ}^{ξ} \frac{d x}{x - α - i β} = \int_{- ξ}^{ξ} \frac{x - α + i β}{{(x - α)}^{2} + β^{2}} d x

= \frac{1}{2} {[l n {{(x - α)}^{2} + β^{2}}]}_{- ξ}^{ξ} + i β \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}}

= \frac{1}{2} l n (\frac{{(ξ - α)}^{2} + β^{2}}{{(ξ + α)}^{2}) + β^{2}}) + i β \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}}

c h a 7 g e m e n t x - α =∣ β ∣ u g i v e \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}}

= \int_{\frac{- ξ - α}{∣ β ∣}}^{\frac{ξ - α}{∣ β ∣}} \frac{∣ β ∣ d u}{β^{2} (1 + u^{2})} = \frac{1}{∣ β ∣} {a r c t a n (\frac{ξ - α}{∣ β ∣}) + a r c t a n (\frac{ξ + α}{∣ β ∣})}

c a s e 1 β > 0 \Rightarrow {l i m}_{ξ \to + \infty} I (ξ) = i β \frac{1}{β} (\frac{π}{2} + \frac{π}{2}) = i π

c a s e 2 β < 0 \Rightarrow {l i m}_{ξ \to + \infty} I (ξ) = - i β \frac{1}{β} (\frac{π}{2} + \frac{π}{2}) = - i π .

Commented by mathmax by abdo last updated on 01/Aug/19

f i n a l l y \int_{- \infty}^{+ \infty} \frac{d x}{x - a} = i π i f i m (a) > 0 a n d - i π i f i m (a) < 0

Commented by mathmax by abdo last updated on 01/Aug/19

2) l e t d e c o m p o s e F (x) = \frac{1}{x^{4} + 1}

x^{4} + 1 = 0 \Rightarrow x^{4} = - 1 \Rightarrow {({r e}^{i θ})}^{4} = e^{i (2 k + 1) π} \Rightarrow r = 1 a n d θ = (2 k + 1) \frac{π}{4}

s o t h e r o o t s a r e z_{k} = e^{i (2 k + 1) \frac{π}{4}} k \in [[0, 3]]

z_{0} = e^{\frac{i 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 (x) = \sum_{i = 0}^{3} \frac{λ_{i}}{x - z_{i}} a n d λ_{i} = \frac{1}{4 z_{i}^{3}} = - \frac{1}{4} z_{i} \Rightarrow

\int_{0}^{+ \infty} \frac{d x}{x^{4} + 1} = \frac{1}{2} \int_{- \infty}^{+ \infty} F (x) d x = \frac{1}{2} \int_{- \infty}^{+ \infty} \sum_{i = 0}^{3} (- \frac{1}{4}) \frac{z_{i}}{x - z_{i}} d x

= - \frac{1}{8} \sum_{i = 0}^{3} \int_{- \infty}^{+ \infty} \frac{z_{i}}{x - z_{i}} d x

= - \frac{1}{8} {z_{0} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{0}} + z_{1} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{1}} + z_{2} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{2}} + z_{3} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{3}}}

= - \frac{1}{8} {i π (z_{0}) + i π z_{1} - i π z_{2} - i π z_{3}}

= - \frac{i π}{8} {z_{0} + z_{1} - (z_{2} + z_{3})} b u t z_{0} + z_{1} = e^{\frac{i π}{4}} + e^{i (π - \frac{π}{4})}

= e^{\frac{i π}{4}} - e^{- \frac{i π}{4}} = 2 i s i n (\frac{π}{4}) = 2 i \frac{\sqrt{2}}{2} = i \sqrt{2}

z_{2} + z_{3} = e^{i \frac{5 π}{4}} + e^{\frac{i 7 π}{4}} = e^{i (π + \frac{π}{4})} + e^{i (2 π - \frac{π}{4})} = - e^{\frac{i π}{4}} + e^{- \frac{i π}{4}}

= - (e^{\frac{i π}{4}} - e^{- \frac{i π}{4}}) = - 2 i s i n (\frac{π}{4}) = - 2 i \frac{\sqrt{2}}{2} = - i \sqrt{2} \Rightarrow

\int_{0}^{\infty} \frac{d x}{1 + x^{4}} = - \frac{i π}{8} {i \sqrt{2} + i \sqrt{2}} = \frac{π}{8} (2 \sqrt{2}) = \frac{π \sqrt{2}}{4}