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Question Number 27620 by abdo imad last updated on 11/Jan/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{{(-1)}^{x^2}}{3+x^2}dx.$$

Commented by abdo imad last updated on 12/Jan/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{{(-1)}^{x^2}}{3+x^2}dxdueto(-1)=e^{i\pi }$$$$I={\int }_0^{\mathrm{\infty }}\frac{e^{i\pi x^2}}{3+x^2}dx=\frac{1}{2}{\int }_R\frac{e^{i\pi x^2}}{3+x^2}dxletintroducethe$$$$complexfunctionf\left(z\right)=\frac{e^{i\pi z^2}}{z^2+3}wehave$$$$f\left(z\right)=\frac{e^{i\pi z^2}}{(z-i\sqrt{3})(z+i\sqrt{3)}}thepolesoffare{z}_1=i\sqrt{3}and$$$$z_2=-i\sqrt{3}theresidustheoremgive$$$${\int }_{R}f\left(z\right)dz=2i\pi Res(f,i\sqrt{3})but$$$$Res(f,i\sqrt{3})={lim}_{z->i\sqrt{3}}(z-i\sqrt{3})f\left(z\right)={lim}_{z->i\sqrt{3}}\frac{e^{i\pi z^2}}{z+i\sqrt{3}}$$$$=\frac{e^{i\pi (-3)}}{2i\sqrt{3}}=\frac{{(-1)}^{-3}}{2i\sqrt{3}}=\frac{-1}{2i\sqrt{3}}$$$${\int }_{R}f\left(z\right)dz=2i\pi .\frac{-1}{2i\sqrt{3}}=\frac{-\pi }{\sqrt{3}}andI=\frac{-\pi }{2\sqrt{3}}.$$

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