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Question Number 36189 by prof Abdo imad last updated on 30/May/18

$$letF\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-x^{2}t}\sqrt{t}}{1+t^2}dt$$$$calculate{lim}_{x\to +\mathrm{\infty }}F\left(x\right).$$$$$$

Commented by math khazana by abdo last updated on 18/Aug/18

$$changement\sqrt{t}=ugive$$$$F\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-x^2{u}^2}u}{1+u^4}\left(2u\right)du$$$$=2{\int }_0^{\mathrm{\infty }}\frac{u^2{e}^{-x^2{u}^2}}{1+u^4}du$$$${=}_{xu=\alpha }2{\int }_0^{\mathrm{\infty }}\frac{{\alpha }^2}{x^2}\frac{e^{-{\alpha }^2}}{1+\frac{{\alpha }^4}{x^4}}\frac{1}{x}d\alpha$$$$=2{\int }_0^{\mathrm{\infty }}\frac{{\alpha }^2{e}^{-{\alpha }^2}}{x^3+\frac{{\alpha }^4}{x}}d\alpha =2x{\int }_0^{\mathrm{\infty }}\frac{{\alpha }^2{e}^{-{\alpha }^2}}{x^4+{\alpha }^4}d\alpha$$$$\Rightarrow F\left(x\right)⩽\frac{2}{x^3}{\int }_0^{\mathrm{\infty }}{\alpha }^2{e}^{-{\alpha }^2}d\alpha but{\int }_0^{\mathrm{\infty }}{\alpha }^2{e}^{-{\alpha }^2}d\alpha$$$$converges\Rightarrow {lim}_{x\to +\mathrm{\infty }}F\left(x\right)=0.$$

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