Question Number 101268 by mathmax by abdo last updated on 01/Jul/20

calculate ∫_(−∞) ^∞ ((cos(arctan(2x+1)))/(x^2 +2x+2))dx

Answered by mathmax by abdo last updated on 02/Jul/20

I =∫_(−∞) ^(+∞) ((cos(arctan(2x+1)))/(x^2 +2x+2))dx changement 2x+1 =t give I =∫_(−∞) ^(+∞) ((cos(arctant))/((((t−1)/2))^2 +2(((t−1)/2))+2)) (dt/2) =(1/2) ∫_(−∞) ^(+∞) ((cos(arctant))/((((t−1)^2 )/4)+t+1)) dt =2 ∫_(−∞) ^(+∞) ((cos(arctant))/((t−1)^2 +4t+4))dt =2 ∫_(−∞) ^(+∞) ((cos(arctant))/(t^2 −2t+1 +4t +4))dt =2 ∫_(−∞) ^(+∞) ((cos(arctant))/(t^2 +2t+5))dt =2 Re(∫_(−∞) ^(+∞) (e^(iarctant) /(t^(2 ) +2t +5))dt) let ϕ(z) =(e^(iarctanz) /(z^2 +2z +5)) poles of ϕ? z^2 +2z +5 =0→Δ^′ =1−5 =−4 ⇒z_1 =−1+2i and z_2 =−1−2i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) ϕ(z) =(e^(iarctan(z)) /((z−z_1 )(z−z_2 ))) ⇒Res(ϕ,z_1 ) =(e^(iarctanz_1 ) /(z_1 −z_2 )) =(e^(iarctan(−1+2i)) /(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ×(e^(iarctan(−1+2i)) /(4i)) =(π/2) e^(iarctan(−1+2i)) we know arctanz =(1/(2i))ln(((1+iz)/(1−iz))) ⇒arctan(−1+2i) =(1/(2i))ln(((1+i(−1+2i))/(1−i(−1+2i)))) =(1/(2i))ln(((1−i−2)/(1+i+2))) =(1/(2i))ln(((−1−i)/(3+i))) also ((−1−i)/(3+i)) =(((√2)e^(i((5π)/4)) )/((√(10))e^(iarctan((1/3))) )) =((√2)/(√(10))) e^(i(((5π)/4)−arctan((1/3)))) ⇒ arctan(−1+2i) =(1/(2i))ln(((√2)/(√(10)))) +(1/(2i))×i(((5π)/4)−arctan((1/3))) =(1/(4i))ln((1/5))+(1/2)(((5π)/4) −arctan((1/3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =(π/2)e^(i((1/(4i))ln((1/5))+((5π)/8)−(1/2)arctan((1/3)))) =(π/2)e^(−((ln5)/4)) { cos(((5π)/8)−(1/2)arctan((1/3))) +isin(....)} ⇒ I =π e^(−((ln5)/4)) cos(((5π)/8)−(1/2)arctan((1/3)))