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Question Number 26057 by abdo imad last updated on 18/Dec/17

$$findaintegralformofL\left(e^{-x^2}\right)$$$$L\left(f\right)meanslaplacetransformoff.$$

Commented by abdo imad last updated on 21/Dec/17

$$wedefineL\left(f\left(x\right)\right)={\int }_0^{\mathrm{\infty }}f\left(t\right)e^{-xt}dtwithx⩾0with{\int }_0^{\mathrm{\infty }}/f\left(t\right)/dtconvergent$$$$L\left(e^{-x^2}\right)={\int }_0^{\mathrm{\infty }}{e}^{-t^2}.e^{-xt}dt={\int }_0^{\mathrm{\infty }}{e}^{-(t^2+xt)}dt={\int }_0^{\mathrm{\infty }}{e}^{-(t^2+2\frac{x}{2}t+\frac{x^2}{4}-\frac{x^2}{4})}dt$$$$=e^{\frac{x^2}{4}}{\int }_0^{\mathrm{\infty }}{e}^{-{(t+\frac{x}{2})}^2}dtandbythechangement\alpha =t+\frac{x}{2}$$$$L\left(e^{-x^2}\right)=e^{\frac{x^2}{4}}{\int }_{\frac{x}{2}}^{\propto }{e}^{-{\alpha }^2}d\alpha =e^{\frac{x^2}{4}}({\int }_{\frac{x}{2}}^0{e}^{-{\alpha }^2}d\alpha +{\int }_0^{\propto }{e}^{-{\alpha }^2}d\alpha )$$$$but{\int }_0^{\propto }{e}^{-{\alpha }^2}d\alpha =\frac{\sqrt{\pi }}{2}$$$$\Rightarrow L\left(e^{-x^2}\right)=e^{\frac{x^2}{4}}(\frac{\sqrt{\pi }}{2}-{\int }_0^{\frac{x}{2}}{e}^{-{\alpha }^2}d\alpha )$$$$$$

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