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find-cos-tx-1-x-2-2-dx-with-t-0-




Question Number 33986 by abdo imad last updated on 28/Apr/18
find ∫_(−∞) ^(+∞)   ((cos(tx))/((1+x^2 )^2 )) dx with t≥0
find+cos(tx)(1+x2)2dxwitht0
Commented by math khazana by abdo last updated on 01/May/18
let introduce the complex function  ϕ(z)= (e^(itz) /((1+z^2 )^2 ))  we have  I =∫_(−∞) ^(+∞)   ((cos(tx))/((1+x^2 )^2 ))dx= Re( ∫_(−∞) ^(+∞)   (e^(itx) /((1+x^2 )^2 ))dx)  ϕ(z)=  (e^(itz) /((z−i)^2 (z+i)^2 )) so the poles of ϕ are  i and−i( doubles)  ∫_(−∞) ^(+∞)   ϕ(z)dz =2iπ Res(ϕ,i)  Res(ϕ,i) =lim_(z→i)   (1/((2−1)!)) ((z−i)^2 f(z))^′   =lim_(z→i)  (  (e^(itz) /((z+i)^2 )))^′   =lim_(z→i)    ((ite^(itz)  (z+i)^2  −2(z+i)e^(itz) )/((z+i)^4 ))  =lim_(z→i)    (((z+i)it e^(itz)  −2 e^(itz) )/((z+i)^3 ))  =(((2i)it e^(−t)    −2 e^(−t) )/((2i)^3 )) = ((−4 e^(−t) )/(−8i)) = (1/(2i)) e^(−t)   ∫_(−∞) ^(+∞)   ϕ(z)dz =2iπ (e^(−t) /(2i)) =π e^(−t)   ⇒ I= π e^(−t)
letintroducethecomplexfunctionφ(z)=eitz(1+z2)2wehaveI=+cos(tx)(1+x2)2dx=Re(+eitx(1+x2)2dx)φ(z)=eitz(zi)2(z+i)2sothepolesofφareiandi(doubles)+φ(z)dz=2iπRes(φ,i)Res(φ,i)=limzi1(21)!((zi)2f(z))=limzi(eitz(z+i)2)=limziiteitz(z+i)22(z+i)eitz(z+i)4=limzi(z+i)iteitz2eitz(z+i)3=(2i)itet2et(2i)3=4et8i=12iet+φ(z)dz=2iπet2i=πetI=πet

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