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Question Number 27620 by abdo imad last updated on 11/Jan/18
find the value of  ∫_0 ^∞  (((−1)^x^2  )/(3+x^2 ))dx .
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo imad last updated on 12/Jan/18
let put I= ∫_0 ^∞  (((−1)^x^2  )/(3+x^2 ))dx   due to (−1)=e^(iπ)   I = ∫_0 ^∞    (e^(iπx^2 ) /(3+x^2 ))dx = (1/2) ∫_R   (e^(iπx^2 ) /(3+x^2 ))dx  let introduce the  complex function  f(z)= (e^(iπz^2 ) /(z^2  +3))  we have  f(z)= (e^(iπz^2 ) /((z −i(√3))(z+i(√(3)))))  the poles of f are z_1 = i(√3) and  z_2 =−i(√3)   the residus theorem give  ∫_R f(z)dz= 2iπ Res(f,i(√3))  but  Res(f,i(√3))= lim_(z−>i(√3)) (z −i(√3))f(z)= lim_(z−>i(√3))   (e^(iπz^2 ) /(z+i(√3)))  =  (e^(iπ(−3)) /(2i(√3)))=  (((−1)^(−3) )/(2i(√3)))= ((−1)/(2i(√3)))  ∫_(R ) f(z)dz= 2iπ.((−1)/(2i(√3)))  = ((−π)/( (√3)))      and   I= ((−π)/(2(√3)))  .
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:\:\:{due}\:{to}\:\left(−\mathrm{1}\right)={e}^{{i}\pi} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{{i}\pi{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{R}} \:\:\frac{{e}^{{i}\pi{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:\:{let}\:{introduce}\:{the} \\ $$$${complex}\:{function}\:\:{f}\left({z}\right)=\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:{we}\:{have} \\ $$$${f}\left({z}\right)=\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}\:−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\left.\mathrm{3}\right)}\right.}\:\:{the}\:{poles}\:{of}\:{f}\:{are}\:{z}_{\mathrm{1}} =\:{i}\sqrt{\mathrm{3}}\:{and} \\ $$$${z}_{\mathrm{2}} =−{i}\sqrt{\mathrm{3}}\:\:\:{the}\:{residus}\:{theorem}\:{give} \\ $$$$\int_{{R}} {f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\:{Res}\left({f},{i}\sqrt{\mathrm{3}}\right)\:\:{but} \\ $$$${Res}\left({f},{i}\sqrt{\mathrm{3}}\right)=\:{lim}_{{z}−>{i}\sqrt{\mathrm{3}}} \left({z}\:−{i}\sqrt{\mathrm{3}}\right){f}\left({z}\right)=\:{lim}_{{z}−>{i}\sqrt{\mathrm{3}}} \:\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{{z}+{i}\sqrt{\mathrm{3}}} \\ $$$$=\:\:\frac{{e}^{{i}\pi\left(−\mathrm{3}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}=\:\:\frac{\left(−\mathrm{1}\right)^{−\mathrm{3}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}=\:\frac{−\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$$\int_{{R}\:} {f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi.\frac{−\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\:=\:\frac{−\pi}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:{and}\:\:\:{I}=\:\frac{−\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:. \\ $$

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