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Question Number 92993 by mathmax by abdo last updated on 10/May/20
calculate ∫_(−∞) ^(+∞) ((cos(cosx−sinx))/(x^2 +4))dx
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({cosx}−{sinx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$
Commented by mathmax by abdo last updated on 10/May/20
I =∫_(−∞) ^(+∞) ((cos(cosx−sinx))/(x^2 +4))dx ⇒ I =∫_(−∞) ^(+∞) ((cos((√2)cos(x+(π/4))))/(x^2 +4))dx =_(x+(π/4)=t) ∫_(−∞) ^(+∞) ((cos((√2)t))/((t−(π/4))^2 +4))dt =Re(∫_(−∞) ^(+∞) (e^(i(√2)t) /((t−(π/4))^2 +4))) let ϕ(z) =(e^(iz(√2)) /((z−(π/4))^2 +4)) poles of ϕ? ϕ(z) =(e^(iz(√2)) /((z−(π/4))^2 −(2i)^2 )) =(e^(iz(√2)) /((z−(π/4)−2i)(z−(π/4)+2i))) so the poles of ϕ are (π/4)+2i and (π/4)−2i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(π/4)+2i) =2iπ(e^(i(√2)((π/4)+2i)) /(4i)) =(π/2) e^((iπ(√2))/4) e^(−2(√2)) =(π/2)e^(−2(√2)) {cos(((π(√2))/4))+isin(((π(√2))/4))} ⇒ I =(π/2)e^(−2(√2)) cos(((π(√2))/4))
$${I}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({cosx}−{sinx}\right)}{{x}^{\mathrm{2}} +\mathrm{4}}{dx}\:\Rightarrow\:{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\sqrt{\mathrm{2}}{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$$$=_{{x}+\frac{\pi}{\mathrm{4}}={t}} \:\:\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\sqrt{\mathrm{2}}{t}\right)}{\left({t}−\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} \:+\mathrm{4}}{dt}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\sqrt{\mathrm{2}}{t}} }{\left({t}−\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} +\mathrm{4}}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{e}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}−\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} +\mathrm{4}}\:\:\:{poles}\:{of}\:\varphi? \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}−\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\mathrm{2}{i}\right)^{\mathrm{2}} }\:=\frac{{e}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}−\frac{\pi}{\mathrm{4}}−\mathrm{2}{i}\right)\left({z}−\frac{\pi}{\mathrm{4}}+\mathrm{2}{i}\right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\frac{\pi}{\mathrm{4}}+\mathrm{2}{i}\:{and}\:\frac{\pi}{\mathrm{4}}−\mathrm{2}{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{\pi}{\mathrm{4}}+\mathrm{2}{i}\right)\:=\mathrm{2}{i}\pi\frac{{e}^{{i}\sqrt{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{i}\right)} }{\mathrm{4}{i}} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{e}^{\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{4}}} \:{e}^{−\mathrm{2}\sqrt{\mathrm{2}}} =\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{2}\sqrt{\mathrm{2}}} \:\:\left\{{cos}\left(\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\right)+{isin}\left(\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{2}\sqrt{\mathrm{2}}} \:{cos}\left(\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\right) \\ $$$$ \\ $$

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