Question Number 100511 by mathmax by abdo last updated on 27/Jun/20

calculate  ∫_(−∞) ^∞   ((arctan(cosx +sinx))/(x^2  +4)) dx

Answered by abdomathmax last updated on 28/Jun/20

I =∫_(−∞) ^(+∞ )  ((arctan(cosx +sinx))/(x^2  +4))dx ⇒  I =∫_(−∞) ^(+∞)  ((arctan((√2)cos(x−(π/4))))/(x^2  +4))dx let  ϕ(z) =((arctan((√2)cos(z−(π/4))))/(z^2 +4)) the poles of  ϕ are  2i and −2i ⇒  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,2i)  =2iπ ×((arctan((√2)cos(2i−(π/4))))/(4i))  =(π/2) arctan((√2) cos(2i−(π/4)))   cos(2i−(π/4)) =((e^(i(2i−(π/4)))  +e^(−i(2i−(π/4))) )/2)  =((e^(−2)  e^(−((iπ)/4))  +e^2  e^((iπ)/4) )/2)  =((e^(−2) ((1/(√2))−(i/(√2)))+e^2 ((1/(√2))+(i/(√2))))/2)  =(1/(2(√2))){  e^(−2) (1−i)+e^2 (1+i)}=....after we use  arctanz =(1/(2i))ln(((1+iz)/(1−iz)))...