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Question Number 31093 by abdo imad last updated on 02/Mar/18

$$calculate{\int }_0^{\mathrm{\infty }}{e}^{-x^2}cos\left(2xy\right)dx.$$

Commented by abdo imad last updated on 04/Mar/18

$$letputF\left(y\right)={\int }_0^{\mathrm{\infty }}{e}^{-x^2}cos\left(2xy\right)dx=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-x^2}cos\left(2xy\right)d3$$$$F\left(y\right)=\frac{1}{2}Re\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2+2ixy}dx\right)but$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2+2ixy}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(x^2+2x\left(iy\right)+{\left(iy\right)}^2-{\left(iy\right)}^2)}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{(x+iy)}^2-y^2}dx=e^{-y^2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}du(ch=x+iy=u)$$$$=\sqrt{\pi }{e}^{-y^2}\Rightarrow F\left(y\right)=\frac{\sqrt{\pi }}{2}{e}^{-y^2}.$$

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