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

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

$$\mathrm{calculate}\:\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\alpha\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}\:\:\left(\alpha\:\mathrm{real}\right) \\ $$

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))) .

$$\mathrm{A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\alpha\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\:\mathrm{A}\:=\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{x}} }{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}\right)\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{z}} }{\mathrm{z}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$$\mathrm{we}\:\mathrm{hsve}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{z}} }{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{i}\right)}\:=\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{z}} }{\left(\mathrm{z}^{\mathrm{2}} \:−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)\left(\mathrm{z}^{\mathrm{2}} \:−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)} \\ $$$$=\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{z}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)}\:\:\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\left\{\:\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)+\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\varphi\left(\mathrm{z}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\:\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } }{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{2isin}\left(\frac{\pi}{\mathrm{2}}\right)\right)} \\ $$$$=\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:×\mathrm{e}^{\mathrm{i}\alpha\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\sqrt{\mathrm{2}}}\right)} }{\mathrm{4i}}\:=\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:×\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:\mathrm{e}^{\frac{\mathrm{i}\alpha}{\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\:=\frac{\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\:\mathrm{e}^{\mathrm{i}\left(−\frac{\pi}{\mathrm{4}}+\frac{\alpha}{\sqrt{\mathrm{2}}}\right)} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\varphi\left(\mathrm{z}\right) \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\:\:\frac{\mathrm{e}^{\mathrm{i}\alpha\:\left(−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} }{−\mathrm{2}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(−\mathrm{2i}\right)}\:=\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{4i}}\:\mathrm{e}^{\mathrm{i}\alpha\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\sqrt{\mathrm{2}}}\right)} \:=\frac{\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\:\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\sqrt{\mathrm{2}}}\right)} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\frac{\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\:\mathrm{e}^{\mathrm{i}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{\alpha}{\sqrt{\mathrm{2}}}\right)} \:+\frac{\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\:\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\sqrt{\mathrm{2}}}\right)} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \left\{\:\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\sqrt{\mathrm{2}}}\right)} \:+\mathrm{e}^{−\mathrm{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\sqrt{\mathrm{2}}}\right)} \right\}\:\:=\frac{\pi}{\mathrm{2}}\:\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:\:\:\:\left(\mathrm{2cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\sqrt{\mathrm{2}}}\right)\right) \\ $$$$=\pi\:\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\:−\frac{\alpha}{\sqrt{\mathrm{2}}}\right)\:\Rightarrow\:\mathrm{A}\:=\pi\:\mathrm{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\sqrt{\mathrm{2}}}\right)\:. \\ $$$$ \\ $$

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