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)))...