calculate by complex method ∫_1 ^(+∞) ((xdx)/(x^4 +1))


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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
$let decompose F(x)=(x/(x^4 +1)) poles of F? x^4 +1 =0 ⇔ x^4 =e^((2k+1)π) ⇒ x_k =e^((i(2k+1)π)/4) and k∈[[0,3]] F(x) =Σ_(k.=0) ^3 (a_k /(x−x_k )) with a_(k ) =(x_k /(4x_k ^3 )) =−(1/4)x_k ^2 =−(1/4)e^((i(2k+1)π)/2) =−(1/4) (i)^(2k+1) =−(1/4)(−1)^k i =(i/4)(−1)^k ⇒ ∫_1 ^(+∞) F(x)dx =(i/4) Σ_(k=0) ^3 (−1)^k ∫_1 ^(+∞) (dx/(x−x_k )) we have x_0 =e^((iπ)/4) , x_1 =e^((i3π)/4) , x_2 =e^(i((5π)/4)) =x_1 ^− , x_3 =e^((i7π)/4) =e^(i(((8π−π)/4))) =x_0 ^− ⇒ ∫_1 ^(+∞) F(x)dx =(i/4){ ∫_1 ^(+∞) (dx/(x−x_0 ))−∫_1 ^(+∞) (dx/(x−x_1 )) +∫_1 ^(+∞) (dx/(x−x_1 ^− )) −∫_1 ^(+∞) (dx/(x−x_0 ^− ))} =(i/4)[ln(((x−x_o )/(x−x_1 )))]_1 ^(+∞) +[ln(((x−x_1 ^− )/(x−x_0 ^− )))]_1 ^(+∞) =(i/4)(−ln(((1−x_0 )/(1−x_1 )))−ln(((1−x_1 ^− )/(1−x_0 ^− )))) =(i/4){ ln(((1−x_1 )/(1−x_0 )))+ln(((1−x_0 ^− )/(1−x_1 ^− )))} rest to finish the calculus$
$l e t d e c o m p o s e F (x) = \frac{x}{x^{4} + 1} p o l e s o f F ?$ $x^{4} + 1 = 0 \Leftrightarrow x^{4} = e^{(2 k + 1) π} \Rightarrow x_{k} = e^{\frac{i (2 k + 1) π}{4}} a n d k \in [[0, 3]]$ $F (x) = \sum_{k . = 0}^{3} \frac{a_{k}}{x - x_{k}} w i t h a_{k} = \frac{x_{k}}{4 x_{k}^{3}} = - \frac{1}{4} x_{k}^{2} = - \frac{1}{4} e^{\frac{i (2 k + 1) π}{2}}$ $= - \frac{1}{4} {(i)}^{2 k + 1} = - \frac{1}{4} {(- 1)}^{k} i = \frac{i}{4} {(- 1)}^{k} \Rightarrow$ $\int_{1}^{+ \infty} F (x) d x = \frac{i}{4} \sum_{k = 0}^{3} {(- 1)}^{k} \int_{1}^{+ \infty} \frac{d x}{x - x_{k}}$ $w e h a v e x_{0} = e^{\frac{i π}{4}}, x_{1} = e^{\frac{i 3 π}{4}}, x_{2} = e^{i \frac{5 π}{4}} = {\bar{x}}_{1}, x_{3} = e^{\frac{i 7 π}{4}} = e^{i (\frac{8 π - π}{4})} = {\bar{x}}_{0}$ $\Rightarrow$ $\int_{1}^{+ \infty} F (x) d x = \frac{i}{4} {\int_{1}^{+ \infty} \frac{d x}{x - x_{0}} - \int_{1}^{+ \infty} \frac{d x}{x - x_{1}} + \int_{1}^{+ \infty} \frac{d x}{x - {\bar{x}}_{1}}$ $- \int_{1}^{+ \infty} \frac{d x}{x - {\bar{x}}_{0}}} = \frac{i}{4} {[l n (\frac{x - x_{o}}{x - x_{1}})]}_{1}^{+ \infty} + {[l n (\frac{x - {\bar{x}}_{1}}{x - {\bar{x}}_{0}})]}_{1}^{+ \infty}$ $= \frac{i}{4} (- l n (\frac{1 - x_{0}}{1 - x_{1}}) - l n (\frac{1 - {\bar{x}}_{1}}{1 - {\bar{x}}_{0}}))$ $= \frac{i}{4} {l n (\frac{1 - x_{1}}{1 - x_{0}}) + l n (\frac{1 - {\bar{x}}_{0}}{1 - {\bar{x}}_{1}})} r e s t t o f i n i s h t h e c a l c u l u s$
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
	$∫_0 ^1 ((xdx)/((x^4 +1)))=∫_1 ^(+∞) ((xdx)/((1+x^4 ))) ∫_1 ^(+∞) ((xdx)/(x^4 +1))=(1/2)∫_0 ^∞ ((xdx)/(x^4 +1)) f(z)=(z/(1+z^4 )),let C_R =[0,R]∪{Re^(iθ) ,θ∈[0,(π/2)]}_(=γ) ∪[iR,0];R>1 ∫_C_R f(z)dz=2iπRes(f,z=e^(i(π/4)) )=2iπ.(1/(4(e^(i(π/2)) )))=(π/2) ∫_γ f(z)dz=∫_0 ^(π/2) Rie^(iθ) f(Re^(iθ) )dθ =iR^2 ∫_0 ^(π/2) (e^(2iθ) /(R^4 e^(4iθ) +1))dθ,∣∫f(z)dz∣≤(π/2).(R^2 /(R^4 −1))→0 as R→∞ ∫_(iR) ^0 f(z)dz=i∫_R ^0 f(it)dz=∫_0 ^R (t/(t^4 +1))dt ⇒lim_(R→∞) ∫_C_R f(z)dz=lim_(R→∞) 2∫_0 ^R ((tdt)/(t^4 +1))=(π/2) ∫_0 ^∞ ((tdt)/(t^4 +1)).=(π/4) ∫_1 ^∞ ((xdx)/(x^4 +1))=(π/8) .$
	$\int_{0}^{1} \frac{x d x}{(x^{4} + 1)} = \int_{1}^{+ \infty} \frac{x d x}{(1 + x^{4})}$ $\int_{1}^{+ \infty} \frac{x d x}{x^{4} + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{x d x}{x^{4} + 1}$ $Missing \left or extra \right$ $\int_{C_{R}} f (z) d z = 2 i π R e s (f, z = e^{i \frac{π}{4}}) = 2 i π . \frac{1}{4 (e^{i \frac{π}{2}})} = \frac{π}{2}$ $\int_{γ} f (z) d z = \int_{0}^{\frac{π}{2}} {R i e}^{i θ} f ({R e}^{i θ}) d θ$ $= {i R}^{2} \int_{0}^{\frac{π}{2}} \frac{e^{2 i θ}}{R^{4} e^{4 i θ} + 1} d θ, ∣ \int f (z) d z ∣⩽ \frac{π}{2} . \frac{R^{2}}{R^{4} - 1} \to 0 a s R \to \infty$ $\int_{i R}^{0} f (z) d z = i \int_{R}^{0} f (i t) d z = \int_{0}^{R} \frac{t}{t^{4} + 1} d t$ $\Rightarrow \lim_{R \to \infty} \int_{C_{R}} f (z) d z = \underset{R \to \infty}{\lim 2} \int_{0}^{R} \frac{t d t}{t^{4} + 1} = \frac{π}{2}$ $\int_{0}^{\infty} \frac{t d t}{t^{4} + 1} . = \frac{π}{4}$ $\int_{1}^{\infty} \frac{x d x}{x^{4} + 1} = \frac{π}{8}$ $.$
	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

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