calculate-by-complex-method-1-xdx-x-4-1-

Question Number 87306 by abdomathmax last updated on 03/Apr/20

c a l c u l a t e b y c o m p l e x m e t h o d \int_{1}^{+ \infty} \frac{x d x}{x^{4} + 1}

Commented by Ar Brandon last updated on 03/Apr/20

M y s u g g e s t i o n

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

= \frac{1}{2} {[a r c t a n (x^{2})]}_{1}^{+ \infty} = \frac{1}{2} (\frac{π}{2}) - \frac{1}{2} (\frac{π}{4}) = \frac{π}{8}

Commented by Ar Brandon last updated on 04/Apr/20

B y c o m p l e x m e t h o d

L e t x^{4} + 1 = 0 \Rightarrow x^{4} = - 1

∣ x^{4} ∣= 1 a n d a r g (x^{4}) = π m o d (2 π) \Rightarrow∣ x ∣= 1 a n d a r g (x) = \frac{π}{4} m o d (\frac{π}{2})

\Rightarrow x_{k} = e^{(2 k + 1) \frac{π}{4} i} k \in Z

\Rightarrow x^{4} + 1 = \underset{k = - 2}{\prod^{1}} (x - x_{k})

\Rightarrow \frac{x}{x^{4} + 1} = \frac{x}{\underset{k = - 2}{\prod^{1}} (x - x_{k})} = \underset{k = - 2}{\sum^{1}} \frac{a_{k}}{(x - x_{k})}

a_{k} = \frac{x_{k}}{4 x_{k}^{3}} i . e \frac{P (x_{k})}{Q^{'} (x_{k})}

\Rightarrow \frac{x}{x^{4} + 1} = \underset{k = - 2}{\sum^{1}} \frac{1}{4 (x_{k}^{2}) (x - x_{k})}

\int_{1}^{+ \infty} \frac{x}{x^{4} + 1} d x = \int_{1}^{+ \infty} \underset{k = - 2}{\sum^{1}} \frac{1}{4 (x_{k}^{2}) (x - x_{k})} d x

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

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

Commented by Ar Brandon last updated on 04/Apr/20

= \frac{l n \infty}{4 i} - \frac{1}{4 i} l n (∣ 1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} ∣) - \frac{l n \infty}{4 i} + \frac{1}{4 i} l n (∣ 1 - \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} ∣)

+ \frac{l n \infty}{4 i} - \frac{1}{4 i} l n (∣ 1 - \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} ∣) - \frac{l n \infty}{4 i} + \frac{1}{4 i} l n (∣ 1 + \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} ∣)

= \frac{1}{4 i} l n (∣ 1 + \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} ∣) - \frac{1}{4 i} l n (∣ 1 - \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} ∣)

+ \frac{1}{4 i} l n (∣ 1 - \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} ∣) - \frac{1}{4 i} l n (∣ 1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} ∣)

Commented by mathmax by abdo last updated on 04/Apr/20

t h a n k s s i r

Commented by mathmax by abdo last updated on 04/Apr/20

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

x^{4} + 1 = 0 \Leftrightarrow x^{4} = e^{(2 k + 1) π} \Rightarrow x_{k} = e^{\frac{i (2 k + 1) π}{4}} a n d k \in [[0, 3]]

F (x) = \sum_{k . = 0}^{3} \frac{a_{k}}{x - x_{k}} w i t h a_{k} = \frac{x_{k}}{4 x_{k}^{3}} = - \frac{1}{4} x_{k}^{2} = - \frac{1}{4} e^{\frac{i (2 k + 1) π}{2}}

= - \frac{1}{4} {(i)}^{2 k + 1} = - \frac{1}{4} {(- 1)}^{k} i = \frac{i}{4} {(- 1)}^{k} \Rightarrow

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

w e h a v e x_{0} = e^{\frac{i π}{4}}, x_{1} = e^{\frac{i 3 π}{4}}, x_{2} = e^{i \frac{5 π}{4}} = {\bar{x}}_{1}, x_{3} = e^{\frac{i 7 π}{4}} = e^{i (\frac{8 π - π}{4})} = {\bar{x}}_{0}

\Rightarrow

\int_{1}^{+ \infty} F (x) d x = \frac{i}{4} {\int_{1}^{+ \infty} \frac{d x}{x - x_{0}} - \int_{1}^{+ \infty} \frac{d x}{x - x_{1}} + \int_{1}^{+ \infty} \frac{d x}{x - {\bar{x}}_{1}}

- \int_{1}^{+ \infty} \frac{d x}{x - {\bar{x}}_{0}}} = \frac{i}{4} {[l n (\frac{x - x_{o}}{x - x_{1}})]}_{1}^{+ \infty} + {[l n (\frac{x - {\bar{x}}_{1}}{x - {\bar{x}}_{0}})]}_{1}^{+ \infty}

= \frac{i}{4} (- l n (\frac{1 - x_{0}}{1 - x_{1}}) - l n (\frac{1 - {\bar{x}}_{1}}{1 - {\bar{x}}_{0}}))

= \frac{i}{4} {l n (\frac{1 - x_{1}}{1 - x_{0}}) + l n (\frac{1 - {\bar{x}}_{0}}{1 - {\bar{x}}_{1}})} r e s t t o f i n i s h t h e c a l c u l u s

Commented by Ar Brandon last updated on 04/Apr/20

I s h o u l d b e t h e o n e t h a n k i n g y o u . H a h a!

I l e a r n e d t h e m e t h o d f r o m y o u . {I t}^{'} s t h e m e t h o d y o u

u s e d w h e n i n t e g r a t i n g \frac{x^{6}}{1 + x^{12}}

Commented by Ar Brandon last updated on 04/Apr/20

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Commented by mind is power last updated on 04/Apr/20

n i c e t h e g o a l o f t h i s F o r u m

i s t o h e l p a n d l e a r n d n e w t r i c k s

p e r s o n a l y i l e a r n d n e w i d e a s i n t h i s f o r u m

s o r r y f o r m y E n g l i s h

Commented by Ar Brandon last updated on 05/Apr/20

A n d I^{'} m d o i n g s a m e h e r e .

Answered by mind is power last updated on 04/Apr/20

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

\int_{1}^{+ \infty} \frac{x d x}{x^{4} + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{x d x}{x^{4} + 1}

Missing \left or extra \right

\int_{C_{R}} f (z) d z = 2 i π R e s (f, z = e^{i \frac{π}{4}}) = 2 i π . \frac{1}{4 (e^{i \frac{π}{2}})} = \frac{π}{2}

\int_{γ} f (z) d z = \int_{0}^{\frac{π}{2}} {R i e}^{i θ} f ({R e}^{i θ}) d θ

= {i R}^{2} \int_{0}^{\frac{π}{2}} \frac{e^{2 i θ}}{R^{4} e^{4 i θ} + 1} d θ, ∣ \int f (z) d z ∣⩽ \frac{π}{2} . \frac{R^{2}}{R^{4} - 1} \to 0 a s R \to \infty

\int_{i R}^{0} f (z) d z = i \int_{R}^{0} f (i t) d z = \int_{0}^{R} \frac{t}{t^{4} + 1} d t

\Rightarrow \lim_{R \to \infty} \int_{C_{R}} f (z) d z = \underset{R \to \infty}{\lim 2} \int_{0}^{R} \frac{t d t}{t^{4} + 1} = \frac{π}{2}

\int_{0}^{\infty} \frac{t d t}{t^{4} + 1} . = \frac{π}{4}

\int_{1}^{\infty} \frac{x d x}{x^{4} + 1} = \frac{π}{8}

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Commented by mathmax by abdo last updated on 04/Apr/20

t h a n k s s i r

Commented by mind is power last updated on 04/Apr/20

w i t h e p l e a s u r s i r