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Question Number 76781 by mathmax by abdo last updated on 30/Dec/19

$$find{\int }_0^{\mathrm{\infty }}\frac{1-e^{-x^2}}{x^2}dx$$

Commented by ~blr237~ last updated on 30/Dec/19

$$\mathrm{let}\mathrm{consider}\mathrm{f}\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{1-{\mathrm{e}}^{-{\mathrm{tx}}^2}}{{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{if}\mathrm{t}⩾0\mathrm{f}\left(\mathrm{t}\right)\mathrm{exist}\mathrm{cause}\mathrm{the}\mathrm{function}$$$$\mathrm{x}\to \frac{1-{\mathrm{e}}^{-{\mathrm{tx}}^2}}{{\mathrm{x}}^2}\mathrm{is}\mathrm{continuous}\mathrm{on}[0,\mathrm{\infty }[\mathrm{with}\mathrm{an}\mathrm{extension}\mathrm{in}0\mathrm{due}\mathrm{to}$$$$\underset{x\to 0}{\lim }\frac{1-{\mathrm{e}}^{-{\mathrm{tx}}^2}}{{\mathrm{x}}^2}=\mathrm{t}.\mathrm{so}\mathrm{Df}={\mathbb{R}}_{+}$$$$\mathrm{let}\mathrm{a}>0,\mathrm{b}>0\mathrm{and}(\mathrm{x},\mathrm{t})\in {\mathrm{D}}_{\mathrm{a},\mathrm{b}}=[0,\mathrm{\infty }[\times [\mathrm{a},\mathrm{b}]$$$$\mid {\mathrm{e}}^{-{\mathrm{tx}}^2}\mid ⩽{\mathrm{e}}^{-{\mathrm{ax}}^2}$$$$\mathrm{we}\mathrm{know}\mathrm{that}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ax}}^2}\mathrm{dx}\mathrm{converges}\mathrm{cause}\mathrm{a}>0$$$$\mathrm{so}(\mathrm{x},\mathrm{t})\to \frac{\partial }{\partial \mathrm{t}}\left(\frac{1-{\mathrm{e}}^{-{\mathrm{tx}}^2}}{{\mathrm{x}}^2}\right)\mathrm{verify}\mathrm{the}\mathrm{convergence}\mathrm{hypothesis}\mathrm{on}{\mathrm{D}}_{\mathrm{a},\mathrm{b}}\mathrm{then}\mathrm{f}\in {\mathrm{C}}^1([\mathrm{a},\mathrm{b}],\mathbb{R})\mathrm{\forall }\mathrm{a}>0\mathrm{b}>0$$$$\mathrm{finally}\mathrm{f}\in {\mathrm{C}}^1({\mathbb{R}}_{+},\mathbb{R})$$$$\mathrm{we}\mathrm{can}\mathrm{write}$$$${\mathrm{f}}^{\prime }\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{tx}}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(\mathrm{x}\sqrt{\mathrm{t}}\right)}^2}\frac{\mathrm{d}\left(\mathrm{x}\sqrt{\mathrm{t}}\right)}{\sqrt{\mathrm{t}}}$$$$=\frac{1}{\sqrt{\mathrm{t}}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}=\sqrt{\frac{\pi }{2\mathrm{t}}}$$$$\mathrm{then}\mathrm{f}\left(\mathrm{t}\right)=\sqrt{2\pi \mathrm{t}}+\mathrm{c}$$$$\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{f}\left(0\right)=0=\mathrm{c}(\mathrm{remember}\mathrm{that}0\in \mathrm{Df})$$$$\mathrm{Finally}\mathrm{f}\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{1-{\mathrm{e}}^{-{\mathrm{tx}}^2}}{{\mathrm{x}}^2}\mathrm{dx}=\sqrt{2\pi \mathrm{t}}$$$$\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{f}\left(1\right)={\int }_0^{\mathrm{\infty }}\frac{1-{\mathrm{e}}^{-{\mathrm{x}}^2}}{{\mathrm{x}}^2}\mathrm{dx}=\sqrt{2\pi }$$

Commented by mathmax by abdo last updated on 10/Jan/20

$$thankssir.$$

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