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Question Number 43057 by maxmathsup by imad last updated on 06/Sep/18
find the value of I = ∫_0 ^∞ ((cos(πx^2 ) −sin(πx^2 ))/((1+x^2 )^2 ))dx
$${find}\:{the}\:{value}\:{of}\:\:\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\pi{x}^{\mathrm{2}} \right)\:−{sin}\left(\pi{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 07/Sep/18
we have I = ∫_0 ^∞ ((cos(πx^2 ))/((1+x^2 )^2 ))dx −∫_0 ^∞ ((sin(πx^2 ))/((1+x^2 )^2 ))dx =H −K we have 2H = ∫_(−∞) ^(+∞) ((cos(πx^2 ))/((1+x^2 )^2 ))dx =Re( ∫_(−∞) ^(+∞) (e^(iπx^2 ) /((1+x^2 )^2 ))dx) let ϕ(z)= (e^(iπz^2 ) /((z^2 +1)^2 )) we have ϕ(z) = (e^(iπz^2 ) /((z−i)^2 (z+i)^2 )) the poles of ϕ are +^− i(doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (e^(iπz^2 ) /((z+i)^2 ))}^((1)) = lim_(z→i) ((2iπz e^(iπz^2 ) (z+i)^2 −2(z+i)e^(iπz^2 ) )/((z+i)^4 )) =lim_(z→i) ((2iπ e^(iπz^2 ) (z+i) −2 e^(iπz^2 ) )/((z+i)^3 )) = ((2iπ e^(−iπ) (2i) −2 e^(−iπ) )/(−8i)) = (((−4π−2)e^(−iπ) )/(−8i)) =(((2π +1)e^(−iπ) )/(4i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(((2π+1)e^(−iπ) )/(4i)) =(((2π +1)e^(−iπ) )/2) =((2π +1)/2)(−1) ⇒H =−(1/4)(2π+1) also ∫_0 ^∞ ((sin(πx^2 ))/((1+x^2 )^2 ))dx = Im ((1/2) ∫_(−∞) ^(+∞) (e^(iπx^2 ) /((1+x^2 )^2 ))) =0 ⇒ I =−(π/2) −(1/4) .
$${we}\:{have}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\pi{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:−\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\pi{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:={H}\:−{K}\:\:{we}\:{have} \\ $$$$\mathrm{2}{H}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\pi{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\pi{x}^{\mathrm{2}} } }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\right)\:{let}\:\varphi\left({z}\right)=\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:{but}\:{Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\:\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:=\:{lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{2}{i}\pi{z}\:{e}^{{i}\pi{z}^{\mathrm{2}} } \left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right){e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{\mathrm{2}{i}\pi\:{e}^{{i}\pi{z}^{\mathrm{2}} } \left({z}+{i}\right)\:−\mathrm{2}\:{e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{3}} }\:=\:\frac{\mathrm{2}{i}\pi\:{e}^{−{i}\pi} \left(\mathrm{2}{i}\right)\:−\mathrm{2}\:{e}^{−{i}\pi} }{−\mathrm{8}{i}} \\ $$$$=\:\frac{\left(−\mathrm{4}\pi−\mathrm{2}\right){e}^{−{i}\pi} }{−\mathrm{8}{i}}\:=\frac{\left(\mathrm{2}\pi\:+\mathrm{1}\right){e}^{−{i}\pi} }{\mathrm{4}{i}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\frac{\left(\mathrm{2}\pi+\mathrm{1}\right){e}^{−{i}\pi} }{\mathrm{4}{i}} \\ $$$$=\frac{\left(\mathrm{2}\pi\:+\mathrm{1}\right){e}^{−{i}\pi} }{\mathrm{2}}\:=\frac{\mathrm{2}\pi\:+\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)\:\Rightarrow{H}\:=−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}\pi+\mathrm{1}\right)\:{also} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left(\pi{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\:{Im}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\pi{x}^{\mathrm{2}} } }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)\:=\mathrm{0}\:\Rightarrow\:{I}\:=−\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$$$ \\ $$

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