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)