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Question Number 140768 by EDWIN88 last updated on 12/May/21

 ∫_(−∞) ^∞ ((x^2 +4)/(x^4 +16)) dx =?

x2+4x4+16dx=?

Answered by Ar Brandon last updated on 12/May/21

I=∫_(−∞) ^∞ ((x^2 +4)/(x^4 +16))dx=2∫_0 ^∞ ((x^2 +4)/(x^4 +16))dx     =2∫_0 ^∞ ((1+(4/x^2 ))/(x^2 +((16)/x^2 )))dx=2∫_0 ^∞ ((1+(4/x^2 ))/((x−(4/x))^2 +8))dx     =2∫_(−∞) ^∞ (du/(u^2 +8))=(2/(2(√2)))[tan^(−1) ((u/(2(√2))))]_(−∞) ^∞ =(1/( (√2)))((π/2)+(π/2))=(π/( (√2)))

I=x2+4x4+16dx=20x2+4x4+16dx=201+4x2x2+16x2dx=201+4x2(x4x)2+8dx=2duu2+8=222[tan1(u22)]=12(π2+π2)=π2

Answered by liberty last updated on 12/May/21

consider I=2∫_0 ^∞  ((x^2 +a^2 )/(x^4 +a^4 )) dx ; a = 2  I=2∫_0 ^∞  ((x^2 (1+(a^2 /x^2 )))/(a^2 x^2 ((x^2 /a^2 )+(a^2 /x^2 )))) dx  I=(2/a^2 ) ∫_0 ^∞  (((1+(a^2 /x^2 )))/(((x^2 /a^2 )+(a^2 /x^2 )))) dx   I= (2/a^2 ) ∫_0 ^∞  (((1+(1/y^2 )))/((y^2 +(1/y^2 )))) (a dy ) ; x=ay  I= (2/a) ∫_0 ^( ∞)  (((1+(1/y^2 )))/((y−(1/y))^2 +((√2))^2 )) dy  I= (2/a) .(1/( (√2))) [arctan (((y−(1/y))/( (√2)))) ]_0 ^∞   I= (2/(a(√2))).π , put a = 2  I= (π/( (√2))) = ((π(√2))/2)

considerI=20x2+a2x4+a4dx;a=2I=20x2(1+a2x2)a2x2(x2a2+a2x2)dxI=2a20(1+a2x2)(x2a2+a2x2)dxI=2a20(1+1y2)(y2+1y2)(ady);x=ayI=2a0(1+1y2)(y1y)2+(2)2dyI=2a.12[arctan(y1y2)]0I=2a2.π,puta=2I=π2=π22

Answered by Mathspace last updated on 12/May/21

let Φ=∫_(−∞) ^(+∞)  ((x^2 +4)/(x^(4 ) +16))dx and  Ψ(z)=((z^2  +4)/(z^4  +16))  poles of Ψ?  Ψ)(z)=((z^2 +4)/((z^2 −4i)(z^2 +4i)))  =  ((z^2 +4)/((z−2e^((iπ)/4) )(z+2e^((iπ)/4) )(z−2e^(−((iπ)/4)) )(z+2e^(−((iπ)/4)) )))  ∫_(−∞) ^(+∞)  Ψ(z)dz=2iπ{ Res(Ψ,2e^((iπ)/4) )+Res(Ψ,−2e^(−((iπ)/4)) )}  we have  Res(Ψ,2e^((iπ)/4) ) =((4i+4)/(4e^((iπ)/4) (8i)))=((1+i)/(8i))e^(−((iπ)/4))   Res(Ψ,−e^(−((iπ)/4)) )=((−4i+4)/(−4e^(−((iπ)/4)) (−8i)))  =((1−i)/(8i))e^((iπ)/4)  ⇒  ∫_(−∞) ^(+∞)  Ψ(z)dz =2iπ{((1+i)/(8i))e^(−((iπ)/4)) +((1−i)/(8i))e^((iπ)/4) }  =(π/4){(1+i)e^(−((iπ)/4))  +(1−i)e^((iπ)/4) }  =(π/4){2Re((1+i)e^(−((iπ)/4)) )}  =(π/2) Re(1+i)e^(−((iπ)/4))  =(π/2)Re((√2))  =((π(√2))/2) ⇒Φ=((π(√2))/2)

letΦ=+x2+4x4+16dxandΨ(z)=z2+4z4+16polesofΨ?Ψ)(z)=z2+4(z24i)(z2+4i)=z2+4(z2eiπ4)(z+2eiπ4)(z2eiπ4)(z+2eiπ4)+Ψ(z)dz=2iπ{Res(Ψ,2eiπ4)+Res(Ψ,2eiπ4)}wehaveRes(Ψ,2eiπ4)=4i+44eiπ4(8i)=1+i8ieiπ4Res(Ψ,eiπ4)=4i+44eiπ4(8i)=1i8ieiπ4+Ψ(z)dz=2iπ{1+i8ieiπ4+1i8ieiπ4}=π4{(1+i)eiπ4+(1i)eiπ4}=π4{2Re((1+i)eiπ4)}=π2Re(1+i)eiπ4=π2Re(2)=π22Φ=π22

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