calculate ∫_(−∞) ^(+∞) ((cos(αx))/(x^4 +1))dx (α real)


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Question Number 98587 by mathmax by abdo last updated on 14/Jun/20

$calculate \int_{- \infty}^{+ \infty} \frac{\cos (α x)}{x^{4} + 1} dx (α real)$
	Answered by mathmax by abdo last updated on 15/Jun/20
	$A =∫_(−∞) ^(+∞) ((cos(αx))/(x^4 +1))dx ⇒ A =Re(∫_(−∞) ^(+∞) (e^(iαx) /(x^4 +1))dx) let ϕ(z) =(e^(iαz) /(z^4 +1)) we hsve ϕ(z) =(e^(iαz) /((z^2 −i)(z^2 +i))) =(e^(iαz) /((z^2 −e^((iπ)/2) )(z^2 −e^(−((iπ)/2)) ))) =(e^(iαz) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} we have Res(ϕ,e^(i(π/4)) ) =lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z) =lim_(z→e^((iπ)/4) ) (e^(iαe^((iπ)/4) ) /(2e^((iπ)/4) (2isin((π/2))))) =((e^(−((iπ)/4)) ×e^(iα((1/(√2))+(i/(√2)))) )/(4i)) =((e^(−((iπ)/4)) ×e^(−(α/(√2))) e^((iα)/(√2)) )/(4i)) =(e^(−(α/(√2))) /(4i)) e^(i(−(π/4)+(α/(√2)))) Res(ϕ,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^((iπ)/4) )ϕ(z) =lim_(z→−e^(−((iπ)/4)) ) (e^(iα (−e^(−((iπ)/4)) )) /(−2 e^(−((iπ)/4)) (−2i))) =(e^((iπ)/4) /(4i)) e^(iα(−(1/(√2))+(i/(√2)))) =(e^(−(α/(√2))) /(4i)) e^(i((π/4)−(α/(√2)))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (e^(−(α/(√2))) /(4i)) e^(i(−(π/4) +(α/(√2)))) +(e^(−(α/(√2))) /(4i)) e^(i((π/4)−(α/(√2)))) } =(π/2) e^(−(α/(√2))) { e^(i((π/4)−(α/(√2)))) +e^(−i((π/4)−(α/(√2)))) } =(π/2) e^(−(α/(√2))) (2cos((π/4)−(α/(√2)))) =π e^(−(α/(√2))) cos((π/4) −(α/(√2))) ⇒ A =π e^(−(α/(√2))) cos((π/4)−(α/(√2))) .$
	$A = \int_{- \infty}^{+ \infty} \frac{\cos (α x)}{x^{4} + 1} dx \Rightarrow A = Re (\int_{- \infty}^{+ \infty} \frac{e^{i α x}}{x^{4} + 1} dx) let φ (z) = \frac{e^{i α z}}{z^{4} + 1}$ $we hsve φ (z) = \frac{e^{i α z}}{(z^{2} - i) (z^{2} + i)} = \frac{e^{i α z}}{(z^{2} - e^{\frac{i π}{2}}) (z^{2} - e^{- \frac{i π}{2}})}$ $= \frac{e^{i α z}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} residus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{4}}) + Res (φ, - e^{- \frac{i π}{4}})} we have$ $Res (φ, e^{i \frac{π}{4}}) = \lim_{z \to e^{\frac{i π}{4}}} (z - e^{\frac{i π}{4}}) φ (z) = \lim_{z \to e^{\frac{i π}{4}}} \frac{e^{i α e^{\frac{i π}{4}}}}{{2 e}^{\frac{i π}{4}} (2 isin (\frac{π}{2}))}$ $= \frac{e^{- \frac{i π}{4}} \times e^{i α (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}}{4 i} = \frac{e^{- \frac{i π}{4}} \times e^{- \frac{α}{\sqrt{2}}} e^{\frac{i α}{\sqrt{2}}}}{4 i} = \frac{e^{- \frac{α}{\sqrt{2}}}}{4 i} e^{i (- \frac{π}{4} + \frac{α}{\sqrt{2}})}$ $Res (φ, - e^{- \frac{i π}{4}}) = \lim_{z \to - e^{- \frac{i π}{4}}} (z + e^{\frac{i π}{4}}) φ (z)$ $= \lim_{z \to - e^{- \frac{i π}{4}}} \frac{e^{i α (- e^{- \frac{i π}{4}})}}{- 2 e^{- \frac{i π}{4}} (- 2 i)} = \frac{e^{\frac{i π}{4}}}{4 i} e^{i α (- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})} = \frac{e^{- \frac{α}{\sqrt{2}}}}{4 i} e^{i (\frac{π}{4} - \frac{α}{\sqrt{2}})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{e^{- \frac{α}{\sqrt{2}}}}{4 i} e^{i (- \frac{π}{4} + \frac{α}{\sqrt{2}})} + \frac{e^{- \frac{α}{\sqrt{2}}}}{4 i} e^{i (\frac{π}{4} - \frac{α}{\sqrt{2}})}}$ $= \frac{π}{2} e^{- \frac{α}{\sqrt{2}}} {e^{i (\frac{π}{4} - \frac{α}{\sqrt{2}})} + e^{- i (\frac{π}{4} - \frac{α}{\sqrt{2}})}} = \frac{π}{2} e^{- \frac{α}{\sqrt{2}}} (2 \cos (\frac{π}{4} - \frac{α}{\sqrt{2}}))$ $= π e^{- \frac{α}{\sqrt{2}}} \cos (\frac{π}{4} - \frac{α}{\sqrt{2}}) \Rightarrow A = π e^{- \frac{α}{\sqrt{2}}} \cos (\frac{π}{4} - \frac{α}{\sqrt{2}}) .$
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