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Question Number 28694 by abdo imad last updated on 29/Jan/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}{e}^{-{tx}^2}cosxdxwitht>0.$$

Commented by abdo imad last updated on 29/Jan/18

$$letputI={\int }_0^{\mathrm{\infty }}{e}^{-{tx}^2}cosxdx$$$$I=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{tx}^2+ix}dxbecause{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{tx}^2}sinxdx=0but$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-({\left(\sqrt{t}x\right)}^2-2\sqrt{t}x\frac{i}{2\sqrt{t}}+{\left(\frac{i}{2\sqrt{t}}\right)}^2-{\left(\frac{i}{2\sqrt{t}}\right)}^2)}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{(\sqrt{t}x-\frac{i}{2\sqrt{t}})}^2-\frac{1}{4t}}dx(ch.\sqrt{t}x-\frac{i}{2\sqrt{t}}=u)$$$$=e^{-\frac{1}{4t}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{t}}=\frac{\sqrt{\pi }}{\sqrt{t}}{e}^{-\frac{1}{4t}}({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}du=\sqrt{\pi })so$$$$I=\frac{\sqrt{\pi }}{2\sqrt{t}}{e}^{-\frac{1}{4t}}.$$

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