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Question Number 69379 by mathmax by abdo last updated on 22/Sep/19
calculste f(a) =∫_(−∞) ^(+∞)    (((−1)^x^2  )/(x^2  +a^2 ))dx   with a>0
$${calculste}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:\:\:{with}\:{a}>\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 23/Sep/19
we have f(a) =∫_(−∞) ^(+∞)  (e^(iπx^2 ) /(x^2  +a^2 ))dx   let ϕ(z)=(e^(iπz^2 ) /(x^2  +a^2 ))  ϕ(z) =(e^(iπz^2 ) /((z−ia)(z+ia)))  residus theorem give  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,ia) =2iπ(e^(iπ(ia)^2 ) /(2ia)) =(π/a) e^(−iπa^2 )   =(π/a)(−1)^a^2   ⇒ f(a) =(π/a)(−1)^a^2
$${we}\:{have}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\pi{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:\:\:{let}\:\varphi\left({z}\right)=\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}−{ia}\right)\left({z}+{ia}\right)}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ia}\right)\:=\mathrm{2}{i}\pi\frac{{e}^{{i}\pi\left({ia}\right)^{\mathrm{2}} } }{\mathrm{2}{ia}}\:=\frac{\pi}{{a}}\:{e}^{−{i}\pi{a}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{{a}}\left(−\mathrm{1}\right)^{{a}^{\mathrm{2}} } \:\Rightarrow\:{f}\left({a}\right)\:=\frac{\pi}{{a}}\left(−\mathrm{1}\right)^{{a}^{\mathrm{2}} } \\ $$

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