Question Number 100746 by Rohit@Thakur last updated on 28/Jun/20

∫_(−∞) ^∞ ((cos3x)/((1+x^2 )^2 ))dx

Answered by mathmax by abdo last updated on 28/Jun/20

A =∫_(−∞) ^(+∞) ((cos(3x))/((x^2 +1)^2 ))dx ⇒ A =Re(∫_(−∞) ^(+∞) (e^(3ix) /((x^2 +1)^2 ))dx) let ϕ(z)=(e^(3iz) /((z^2 +1)^2 )) we have ∣ϕ(z)z∣ →0 (z→∞) and ϕ(z) =(e^(3iz) /((z−i)^2 (z+i)^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(e^(i3z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((3ie^(i3z) (z+i)^2 −2(z+i)e^(i3z) )/((z+i)^4 )) =lim_(z→i) ((3ie^(i3z) (z+i)−2 e^(i3z) )/((z+i)^3 )) =lim_(z→i) (((3i(z+i)−2)e^(i3z) )/((z+i)^3 )) =(((3i(2i)−2)e^(i(3i)) )/((2i)^3 )) =(((−6−2)e^(−3) )/(−8i)) =(e^(−3) /i) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(e^(−3) /i) =2π e^(−3) ⇒ A =((2π)/e^3 )

Commented byRohit@Thakur last updated on 28/Jun/20

thanks sir

Commented byRohit@Thakur last updated on 28/Jun/20

sir if you can plz provide the solution by real analysis..then it will be very beneficial for me

Commented byabdomathmax last updated on 28/Jun/20

impossible to solve this kind of integral with elementary functions...