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


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

$calculate \int_{- \infty}^{\infty} \frac{\cos (\arctan (2 x + 1))}{x^{2} + 2 x + 2} dx$
	Answered by mathmax by abdo last updated on 02/Jul/20
	$I =∫_(−∞) ^(+∞) ((cos(arctan(2x+1)))/(x^2 +2x+2))dx changement 2x+1 =t give I =∫_(−∞) ^(+∞) ((cos(arctant))/((((t−1)/2))^2 +2(((t−1)/2))+2)) (dt/2) =(1/2) ∫_(−∞) ^(+∞) ((cos(arctant))/((((t−1)^2 )/4)+t+1)) dt =2 ∫_(−∞) ^(+∞) ((cos(arctant))/((t−1)^2 +4t+4))dt =2 ∫_(−∞) ^(+∞) ((cos(arctant))/(t^2 −2t+1 +4t +4))dt =2 ∫_(−∞) ^(+∞) ((cos(arctant))/(t^2 +2t+5))dt =2 Re(∫_(−∞) ^(+∞) (e^(iarctant) /(t^(2 ) +2t +5))dt) let ϕ(z) =(e^(iarctanz) /(z^2 +2z +5)) poles of ϕ? z^2 +2z +5 =0→Δ^′ =1−5 =−4 ⇒z_1 =−1+2i and z_2 =−1−2i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) ϕ(z) =(e^(iarctan(z)) /((z−z_1 )(z−z_2 ))) ⇒Res(ϕ,z_1 ) =(e^(iarctanz_1 ) /(z_1 −z_2 )) =(e^(iarctan(−1+2i)) /(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ×(e^(iarctan(−1+2i)) /(4i)) =(π/2) e^(iarctan(−1+2i)) we know arctanz =(1/(2i))ln(((1+iz)/(1−iz))) ⇒arctan(−1+2i) =(1/(2i))ln(((1+i(−1+2i))/(1−i(−1+2i)))) =(1/(2i))ln(((1−i−2)/(1+i+2))) =(1/(2i))ln(((−1−i)/(3+i))) also ((−1−i)/(3+i)) =(((√2)e^(i((5π)/4)) )/((√(10))e^(iarctan((1/3))) )) =((√2)/(√(10))) e^(i(((5π)/4)−arctan((1/3)))) ⇒ arctan(−1+2i) =(1/(2i))ln(((√2)/(√(10)))) +(1/(2i))×i(((5π)/4)−arctan((1/3))) =(1/(4i))ln((1/5))+(1/2)(((5π)/4) −arctan((1/3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =(π/2)e^(i((1/(4i))ln((1/5))+((5π)/8)−(1/2)arctan((1/3)))) =(π/2)e^(−((ln5)/4)) { cos(((5π)/8)−(1/2)arctan((1/3))) +isin(....)} ⇒ I =π e^(−((ln5)/4)) cos(((5π)/8)−(1/2)arctan((1/3)))$
	$I = \int_{- \infty}^{+ \infty} \frac{\cos (\arctan (2 x + 1))}{x^{2} + 2 x + 2} dx changement 2 x + 1 = t give$ $I = \int_{- \infty}^{+ \infty} \frac{\cos (arctant)}{{(\frac{t - 1}{2})}^{2} + 2 (\frac{t - 1}{2}) + 2} \frac{dt}{2}$ $= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{\cos (arctant)}{\frac{{(t - 1)}^{2}}{4} + t + 1} dt = 2 \int_{- \infty}^{+ \infty} \frac{\cos (arctant)}{{(t - 1)}^{2} + 4 t + 4} dt$ $= 2 \int_{- \infty}^{+ \infty} \frac{\cos (arctant)}{t^{2} - 2 t + 1 + 4 t + 4} dt = 2 \int_{- \infty}^{+ \infty} \frac{\cos (arctant)}{t^{2} + 2 t + 5} dt$ $= 2 Re (\int_{- \infty}^{+ \infty} \frac{e^{iarctant}}{t^{2} + 2 t + 5} dt) let φ (z) = \frac{e^{iarctanz}}{z^{2} + 2 z + 5} poles of φ ?$ $z^{2} + 2 z + 5 = 0 \to Δ^{^{'}} = 1 - 5 = - 4 \Rightarrow z_{1} = - 1 + 2 i and z_{2} = - 1 - 2 i$ $residus theorem give \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1})$ $φ (z) = \frac{e^{iarctan (z)}}{(z - z_{1}) (z - z_{2})} \Rightarrow Res (φ, z_{1}) = \frac{e^{{iarctanz}_{1}}}{z_{1} - z_{2}} = \frac{e^{iarctan (- 1 + 2 i)}}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{e^{iarctan (- 1 + 2 i)}}{4 i} = \frac{π}{2} e^{iarctan (- 1 + 2 i)}$ $we know arctanz = \frac{1}{2 i} \ln (\frac{1 + iz}{1 - iz}) \Rightarrow \arctan (- 1 + 2 i) = \frac{1}{2 i} \ln (\frac{1 + i (- 1 + 2 i)}{1 - i (- 1 + 2 i)})$ $= \frac{1}{2 i} \ln (\frac{1 - i - 2}{1 + i + 2}) = \frac{1}{2 i} \ln (\frac{- 1 - i}{3 + i}) also$ $\frac{- 1 - i}{3 + i} = \frac{\sqrt{2} e^{i \frac{5 π}{4}}}{\sqrt{10} e^{iarctan (\frac{1}{3})}} = \frac{\sqrt{2}}{\sqrt{10}} e^{i (\frac{5 π}{4} - \arctan (\frac{1}{3}))} \Rightarrow$ $\arctan (- 1 + 2 i) = \frac{1}{2 i} \ln (\frac{\sqrt{2}}{\sqrt{10}}) + \frac{1}{2 i} \times i (\frac{5 π}{4} - \arctan (\frac{1}{3}))$ $= \frac{1}{4 i} \ln (\frac{1}{5}) + \frac{1}{2} (\frac{5 π}{4} - \arctan (\frac{1}{3})) \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = \frac{π}{2} e^{i (\frac{1}{4 i} \ln (\frac{1}{5}) + \frac{5 π}{8} - \frac{1}{2} \arctan (\frac{1}{3}))}$ $= \frac{π}{2} e^{- \frac{\ln 5}{4}} {\cos (\frac{5 π}{8} - \frac{1}{2} \arctan (\frac{1}{3})) + isin (. . . .)} \Rightarrow$ $I = π e^{- \frac{\ln 5}{4}} \cos (\frac{5 π}{8} - \frac{1}{2} \arctan (\frac{1}{3}))$
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