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Question Number 140768 by EDWIN88 last updated on 12/May/21
∫−∞∞x2+4x4+16dx=?
Answered by Ar Brandon last updated on 12/May/21
I=∫−∞∞x2+4x4+16dx=2∫0∞x2+4x4+16dx=2∫0∞1+4x2x2+16x2dx=2∫0∞1+4x2(x−4x)2+8dx=2∫−∞∞duu2+8=222[tan−1(u22)]−∞∞=12(π2+π2)=π2
Answered by liberty last updated on 12/May/21
considerI=2∫∞0x2+a2x4+a4dx;a=2I=2∫∞0x2(1+a2x2)a2x2(x2a2+a2x2)dxI=2a2∫∞0(1+a2x2)(x2a2+a2x2)dxI=2a2∫∞0(1+1y2)(y2+1y2)(ady);x=ayI=2a∫∞0(1+1y2)(y−1y)2+(2)2dyI=2a.12[arctan(y−1y2)]0∞I=2a2.π,puta=2I=π2=π22
Answered by Mathspace last updated on 12/May/21
letΦ=∫−∞+∞x2+4x4+16dxandΨ(z)=z2+4z4+16polesofΨ?Ψ)(z)=z2+4(z2−4i)(z2+4i)=z2+4(z−2eiπ4)(z+2eiπ4)(z−2e−iπ4)(z+2e−iπ4)∫−∞+∞Ψ(z)dz=2iπ{Res(Ψ,2eiπ4)+Res(Ψ,−2e−iπ4)}wehaveRes(Ψ,2eiπ4)=4i+44eiπ4(8i)=1+i8ie−iπ4Res(Ψ,−e−iπ4)=−4i+4−4e−iπ4(−8i)=1−i8ieiπ4⇒∫−∞+∞Ψ(z)dz=2iπ{1+i8ie−iπ4+1−i8ieiπ4}=π4{(1+i)e−iπ4+(1−i)eiπ4}=π4{2Re((1+i)e−iπ4)}=π2Re(1+i)e−iπ4=π2Re(2)=π22⇒Φ=π22
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