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Question Number 194462 by Erico last updated on 08/Jul/23

$$\mathrm{Prove}\mathrm{that}{\underset{0}{\int }}^{+{\mathrm{\infty }}^{}}\frac{1-{\mathit{e}}^{-{\mathit{x}}^2}}{{\mathit{x}}^2}\mathit{d}\mathit{x}=\sqrt{\mathit{\pi }}$$

Answered by mnjuly1970 last updated on 08/Jul/23

$$I={\int }_0^{\mathrm{\infty }}\frac{1-e^{-x^2}}{x^2}dx\stackrel{?}{=}\sqrt{\pi }$$$$I\stackrel{i.b.p}{=}{[-\frac{1}{x}(1-e^{-x^2})]}_0^{\mathrm{\infty }}+2{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx$$$$\underset{integral}{\stackrel{Gauss}{=}}2\left(\frac{\sqrt{\pi }}{2}\right)=\sqrt{\pi }$$$$$$$$$$

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