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Question Number 73333 by mathmax by abdo last updated on 10/Nov/19
calculate ∫_0 ^∞ ((cos(π +2x^2 ))/((x^2 +4)^2 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\pi\:+\mathrm{2}{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
let A =∫_0 ^∞ ((cos(π+2x^2 ))/((x^2 +4)^2 ))dx ⇒−2A = ∫_(−∞) ^(+∞) ((cos(2x^2 ))/((x^2 +4)^2 ))dx =_(x=2t) ∫_(−∞) ^(+∞) ((cos(8t^2 ))/(16(t^2 +1)^2 ))(2dt) =(1/8) ∫_(−∞) ^(+∞) ((cos(8t^2 ))/((t^2 +1)^2 ))dt ⇒ A =−(1/(16)) ∫_(−∞) ^(+∞) ((cos(8t^2 ))/((t^2 +1)^2 ))dt =−(1/(16)) Re(∫_(−∞) ^(+∞) (e^(8it^2 ) /((t^2 +1)^2 ))dt) let ϕ(z)=(e^(8iz^2 ) /((z^2 +1)^2 )) ⇒ϕ(z)=(e^(8iz^2 ) /((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(e^(i8z^2 ) /((z+i)^2 ))}^((1)) =lim_(z→i) ((16iz e^(i8z^2 ) (z+i)^2 −2(z+i)e^(8iz^2 ) )/((z+i)^4 )) =lim_(z→i) (({16iz(z+i)−2}e^(8iz^2 ) )/((z+i)^3 )) =(((−16(2i)−2}e^(−8i) )/(−8i)) =(((16i+1)e^(−8i) )/(4i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(((16i+1)e^(−8i) )/(4i)) =(π/2)(1+16i)( cos(8)−isin(8)) =(π/2){cos(8)−isin(8)+16cos(8)i+16 sin(8)} ⇒ A =−(π/(32))( cos(8) +16 sin(8)}
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\pi+\mathrm{2}{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx}\:\Rightarrow−\mathrm{2}{A}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$$$=_{{x}=\mathrm{2}{t}} \:\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{8}{t}^{\mathrm{2}} \right)}{\mathrm{16}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{2}{dt}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\mathrm{8}{t}^{\mathrm{2}} \right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$${A}\:=−\frac{\mathrm{1}}{\mathrm{16}}\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\mathrm{8}{t}^{\mathrm{2}} \right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:=−\frac{\mathrm{1}}{\mathrm{16}}\:{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{\mathrm{8}{it}^{\mathrm{2}} } }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right) \\ $$$${let}\:\varphi\left({z}\right)=\frac{{e}^{\mathrm{8}{iz}^{\mathrm{2}} } }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{\mathrm{8}{iz}^{\mathrm{2}} } }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\:\left\{\frac{{e}^{{i}\mathrm{8}{z}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{16}{iz}\:{e}^{{i}\mathrm{8}{z}^{\mathrm{2}} } \:\left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){e}^{\mathrm{8}{iz}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left\{\mathrm{16}{iz}\left({z}+{i}\right)−\mathrm{2}\right\}{e}^{\mathrm{8}{iz}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{3}} }\:\:\:=\frac{\left(−\mathrm{16}\left(\mathrm{2}{i}\right)−\mathrm{2}\right\}{e}^{−\mathrm{8}{i}} }{−\mathrm{8}{i}} \\ $$$$=\frac{\left(\mathrm{16}{i}+\mathrm{1}\right){e}^{−\mathrm{8}{i}} }{\mathrm{4}{i}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\left(\mathrm{16}{i}+\mathrm{1}\right){e}^{−\mathrm{8}{i}} }{\mathrm{4}{i}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\mathrm{16}{i}\right)\left(\:{cos}\left(\mathrm{8}\right)−{isin}\left(\mathrm{8}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{{cos}\left(\mathrm{8}\right)−{isin}\left(\mathrm{8}\right)+\mathrm{16}{cos}\left(\mathrm{8}\right){i}+\mathrm{16}\:{sin}\left(\mathrm{8}\right)\right\}\:\Rightarrow \\ $$$${A}\:=−\frac{\pi}{\mathrm{32}}\left(\:{cos}\left(\mathrm{8}\right)\:+\mathrm{16}\:{sin}\left(\mathrm{8}\right)\right\} \\ $$

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