Solve by resideo theorem ∫_(−∞) ^( ∞) (z^2 /(z^4 +1)) dz


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Question Number 164991 by mkam last updated on 24/Jan/22

$S o l v e b y r e s i d e o t h e o r e m \int_{- \infty}^{\infty} \frac{z^{2}}{z^{4} + 1} d z$
Commented by mkam last updated on 24/Jan/22

$? ? ? ? ?$
	Answered by aleks041103 last updated on 26/Jan/22
	$Define the region Ω_r :={z∣Im(z)≥0 and ∣z∣≤r}⊂C where r∈R^+ . Therefore the boundary is: ∂Ω_r =Γ_r ∪Λ_r , where Γ_r :={z∣Im(z)>0 and ∣z∣=r} and Λ_r :={z∣z∈R and z∈[−r,r]}. Then: ∮_( ∂Ω_r ) (z^2 /(z^4 +1))dz=∫_Γ_r (z^2 /(z^4 +1))dz + ∫_Λ_r (z^2 /(z^4 +1))dz= =∫_0 ^π ((r^2 e^(2it) )/(r^4 e^(4it) +1))d(re^(it) )+∫_(−r) ^( r) (z^2 /(z^4 +1))dz if r→∞: lim_(r→∞) ∫_Γ_r (z^2 /(z^4 +1))dz=∫_0 ^π (lim_(r→∞) ((ir^3 e^(3it) )/(r^4 e^(4it) +1)))dt=0 ⇒lim_(r→∞) ∮_( ∂Ω_r ) (z^2 /(z^4 +1))dz=∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz From the residue theorem: lim_(r→∞) ∮_( ∂Ω_r ) (z^2 /(z^4 +1))dz=2πiΣ_j Res((z^2 /(z^4 +1)),z_j ∈Ω_∞ ) Ω_∞ :={z∣Im(z)≥0} ⇒∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz=2πiΣ_j Res((z^2 /(z^4 +1)),Im(z_j )≥0) The poles of (z^2 /(z^4 +1)) are: z^4 +1=0⇒z=e^(iπ/4) ,e^(3iπ/4) ,e^(5iπ/4) ,e^(7iπ/4) or z=(1/( (√2)))+(1/( (√2)))i,−(1/( (√2)))+(1/( (√2)))i,−(1/( (√2)))−(1/( (√2)))i,(1/( (√2)))−(1/( (√2)))i Only the poles at z=e^(iπ/4) and z=e^(3iπ/4) lie inside the region Ω_∞ . ⇒∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz=2πi(Res(e^(iπ/4) )+Res(e^(3iπ/4) )) (z^2 /(z^4 +1))=(z^2 /((z−e^(iπ/4) )(z−e^(3iπ/4) )(z−e^(5iπ/4) )(z−e^(7iπ/4) ))) Therefore the poles are simple ⇒Res(e^(iπ/4) )=lim_(z→e^(iπ/4) ) (((z−e^(iπ/4) )z^2 )/(z^4 +1))=^(l′hopital) =lim_(z→e^(iπ/4) ) ((3z^2 −2ze^(iπ/4) )/(4z^3 ))=(1/4)e^(−iπ/4) Res(e^(3iπ/4) )=lim_(z→e^(3iπ/4) ) (((z−e^(3iπ/4) )z^2 )/(z^4 +1))= =lim_(z→e^(3iπ/4) ) ((3z^2 −2ze^(3iπ/4) )/(4z^3 ))=(1/4)e^(−3iπ/4) ∫_(−∞) ^( ∞) (z^2 /(z^4 +1))dz=2πi((1/4)e^(−iπ/4) +(1/4)e^(−3iπ/4) )= =((πi)/2)((1/( (√2)))−(i/( (√2)))+(−(1/( (√2))))−(i/( (√2))))= =((πi)/2)(−((2i)/( (√2))))= =(π/( (√2)))$
	$D e f i n e t h e r e g i o n$ $Ω_{r} := {z ∣ I m (z) ⩾ 0 a n d ∣ z ∣⩽ r} \subset C$ $w h e r e r \in R^{+} .$ $T h e r e f o r e t h e b o u n d a r y i s :$ $\partial Ω_{r} = Γ_{r} \cup Λ_{r}, w h e r e$ $Γ_{r} := {z ∣ I m (z) > 0 a n d ∣ z ∣= r}$ $a n d Λ_{r} := {z ∣ z \in R a n d z \in [- r, r]} .$ $T h e n :$ $\oint_{\partial Ω_{r}} \frac{z^{2}}{z^{4} + 1} d z = \int_{Γ_{r}} \frac{z^{2}}{z^{4} + 1} d z + \int_{Λ_{r}} \frac{z^{2}}{z^{4} + 1} d z =$ $= \int_{0}^{π} \frac{r^{2} e^{2 i t}}{r^{4} e^{4 i t} + 1} d ({r e}^{i t}) + \int_{- r}^{r} \frac{z^{2}}{z^{4} + 1} d z$ $i f r \to \infty :$ $\underset{r \to \infty}{l i m} \int_{Γ_{r}} \frac{z^{2}}{z^{4} + 1} d z = \int_{0}^{π} (\underset{r \to \infty}{l i m} \frac{{i r}^{3} e^{3 i t}}{r^{4} e^{4 i t} + 1}) d t = 0$ $\Rightarrow \underset{r \to \infty}{l i m} \oint_{\partial Ω_{r}} \frac{z^{2}}{z^{4} + 1} d z = \int_{- \infty}^{\infty} \frac{z^{2}}{z^{4} + 1} d z$ $F r o m t h e r e s i d u e t h e o r e m :$ $\underset{r \to \infty}{l i m} \oint_{\partial Ω_{r}} \frac{z^{2}}{z^{4} + 1} d z = 2 π i \sum_{j} R e s (\frac{z^{2}}{z^{4} + 1}, z_{j} \in Ω_{\infty})$ $Ω_{\infty} := {z ∣ I m (z) ⩾ 0}$ $\Rightarrow \int_{- \infty}^{\infty} \frac{z^{2}}{z^{4} + 1} d z = 2 π i \sum_{j} R e s (\frac{z^{2}}{z^{4} + 1}, I m (z_{j}) ⩾ 0)$ $T h e p o l e s o f \frac{z^{2}}{z^{4} + 1} a r e :$ $z^{4} + 1 = 0 \Rightarrow z = e^{i π / 4}, e^{3 i π / 4}, e^{5 i π / 4}, e^{7 i π / 4}$ $o r z = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i, - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i, - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i, \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i$ $O n l y t h e p o l e s a t$ $z = e^{i π / 4} a n d z = e^{3 i π / 4} l i e i n s i d e t h e r e g i o n$ $Ω_{\infty} .$ $\Rightarrow \int_{- \infty}^{\infty} \frac{z^{2}}{z^{4} + 1} d z = 2 π i (R e s (e^{i π / 4}) + R e s (e^{3 i π / 4}))$ $\frac{z^{2}}{z^{4} + 1} = \frac{z^{2}}{(z - e^{i π / 4}) (z - e^{3 i π / 4}) (z - e^{5 i π / 4}) (z - e^{7 i π / 4})}$ $T h e r e f o r e t h e p o l e s a r e s i m p l e$ $\Rightarrow R e s (e^{i π / 4}) = \underset{z \to e^{i π / 4}}{l i m} \frac{(z - e^{i π / 4}) z^{2}}{z^{4} + 1} \overset{l^{'} h o p i t a l}{=}$ $= \underset{z \to e^{i π / 4}}{l i m} \frac{3 z^{2} - 2 {z e}^{i π / 4}}{4 z^{3}} = \frac{1}{4} e^{- i π / 4}$ $R e s (e^{3 i π / 4}) = \underset{z \to e^{3 i π / 4}}{l i m} \frac{(z - e^{3 i π / 4}) z^{2}}{z^{4} + 1} =$ $= \underset{z \to e^{3 i π / 4}}{l i m} \frac{3 z^{2} - 2 {z e}^{3 i π / 4}}{4 z^{3}} = \frac{1}{4} e^{- 3 i π / 4}$ $\int_{- \infty}^{\infty} \frac{z^{2}}{z^{4} + 1} d z = 2 π i (\frac{1}{4} e^{- i π / 4} + \frac{1}{4} e^{- 3 i π / 4}) =$ $= \frac{π i}{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} + (- \frac{1}{\sqrt{2}}) - \frac{i}{\sqrt{2}}) =$ $= \frac{π i}{2} (- \frac{2 i}{\sqrt{2}}) =$ $= \frac{π}{\sqrt{2}}$
	Commented by Tawa11 last updated on 25/Jan/22

	$Great sir$
	Answered by Ar Brandon last updated on 26/Jan/22

	$I = \int_{- \infty}^{+ \infty} \frac{z^{2}}{z^{4} + 1} d z = \int_{0}^{+ \infty} \frac{(z^{2} + 1) + (z^{2} - 1)}{z^{4} + 1} d z$ $= \int_{0}^{+ \infty} \frac{z^{2} + 1}{z^{4} + 1} d z + \int_{0}^{+ \infty} \frac{z^{2} - 1}{z^{4} + 1} d z$ $= \int_{0}^{+ \infty} \frac{1 + \frac{1}{z^{2}}}{z^{2} + \frac{1}{z^{2}}} d z + \int_{0}^{+ \infty} \frac{1 - \frac{1}{z^{2}}}{z^{2} + \frac{1}{z^{2}}} d z$ $= \int_{0}^{+ \infty} \frac{1 + \frac{1}{z^{2}}}{{(z - \frac{1}{z})}^{2} + 2} d z + \int_{0}^{+ \infty} \frac{1 - \frac{1}{z^{2}}}{{(z + \frac{1}{z})}^{2} - 2} d z$ $= \frac{1}{\sqrt{2}} {[\arctan (\frac{z^{2} - 1}{\sqrt{2} z})]}_{0}^{+ \infty} - \frac{1}{2 \sqrt{2}} {[\ln ∣ \frac{z^{2} + \sqrt{2} z + 1}{z^{2} - \sqrt{2} z + 1} ∣]}_{0}^{+ \infty}$ $= \frac{1}{\sqrt{2}} (\frac{π}{2} + \frac{π}{2}) = \frac{π}{\sqrt{2}}$
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