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Question Number 38469 by maxmathsup by imad last updated on 25/Jun/18

calculate  f(a) = ∫_(−∞) ^(+∞)    ((sin(ax))/(x^2  +x+1))dx  2) find the value of  ∫_(−∞) ^(+∞)    ((sin(3x))/(x^2  +x+1))dx

calculatef(a)=+sin(ax)x2+x+1dx2)findthevalueof+sin(3x)x2+x+1dx

Commented by math khazana by abdo last updated on 28/Jun/18

1) f(a)= Im( ∫_(−∞) ^(+∞)   (e^(iax) /(x^2  +x+1))dx) let consider  the complex function ϕ(z)= (e^(iaz) /(z^2  +z +1))  the poles of ϕ are j=e^(i((2π)/3))  and j^− =e^(−((i2π)/3))   j=−(1/2) +i((√3)/2)  ∫_(−∞) ^(+∞)   ϕ(z)dz =2iπ Res(ϕ,j) but   ϕ(z)=(e^(iaz) /((z−j)(z−j^− ))) ⇒Res(ϕ,j)= (e^(iaj) /(j−j^− ))  = (e^(ia(−(1/2) +((i(√3))/2))) /(2i((√3)/2))) = ((e^(−((a(√3))/2)) {cos((a/2))−isin((a/2))})/(i(√3))) ⇒  ∫_(−∞) ^(+∞)   ϕ(z)dz =2iπ ((e^(−((a(√3))/2)) { cos((a/2))−i sin((a/2))})/(i(√3)))  =((2π)/(√3)) e^(−((a(√3))/2)) { cos((a/2))−i sin((a/2))}⇒  f(a)=Im( ∫_(−∞) ^(+∞)  ϕ(z)dz)=−((2π)/3) e^(−a((√3)/2))  sin((a/2))  2) ∫_(−∞) ^(+∞)   ((sin(3x))/(x^2  +x+1))dx=f(3)=−((2π)/3)e^(−((3(√3))/2))  sin((3/2)).

1)f(a)=Im(+eiaxx2+x+1dx)letconsiderthecomplexfunctionφ(z)=eiazz2+z+1thepolesofφarej=ei2π3andj=ei2π3j=12+i32+φ(z)dz=2iπRes(φ,j)butφ(z)=eiaz(zj)(zj)Res(φ,j)=eiajjj=eia(12+i32)2i32=ea32{cos(a2)isin(a2)}i3+φ(z)dz=2iπea32{cos(a2)isin(a2)}i3=2π3ea32{cos(a2)isin(a2)}f(a)=Im(+φ(z)dz)=2π3ea32sin(a2)2)+sin(3x)x2+x+1dx=f(3)=2π3e332sin(32).

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