Question Number 101970 by Rohit@Thakur last updated on 05/Jul/20

∫_0 ^∞  ((Cos(ax))/(x^2 +b^2 )) dx

Answered by mathmax by abdo last updated on 06/Jul/20

let I =∫_0 ^∞  ((cos(ax))/(x^2  +b^2 ))dx for b>0  we do the changement x =bt ⇒  I =∫_0 ^∞  ((cos(abt))/(b^2 (1+t^2 )))bdt =(1/b)∫_0 ^∞   ((cos(abt))/(t^2  +1))dt   =(1/(2b))∫_(−∞) ^(+∞)  ((cos(abt))/(t^2  +1))dt  =(1/(2b)) Re(∫_(−∞) ^(+∞)  (e^(iabt) /(t^2  +1))dt)  let ϕ(z) =(e^(iabz) /(z^2  +1)) ⇒ϕ(z) =(e^(iabz) /((z−i)(z+i)))  residus tbeorem give ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,i)  =2iπ×(e^(iab(i)) /(2i)) =π e^(−ab)  ⇒ I =(π/(2b)) e^(−ab)