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Question Number 95919 by Abdulrahman last updated on 28/May/20

$$$$$$\int 3^{-{4\mathrm{x}}^2}\mathrm{dx}=?(0,\mathrm{\infty })$$

Answered by mathmax by abdo last updated on 28/May/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{3}^{-{4\mathrm{x}}^2}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{4\mathrm{x}}^2\ln \left(3\right)}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(2\mathrm{x}\sqrt{\ln 3}\right)}^2}\mathrm{dx}$$$${=}_{2\mathrm{x}\sqrt{\ln 3}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\frac{\mathrm{dt}}{2\sqrt{\ln 3}}=\frac{1}{2\sqrt{\ln 3}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}=\frac{1}{2\sqrt{\ln 3}}\times \frac{\sqrt{\pi }}{2}\Rightarrow$$$$\mathrm{I}=\frac{\sqrt{\pi }}{4\sqrt{\ln 3}}.$$

Commented by Abdulrahman last updated on 28/May/20

$$\mathrm{why}\ln \left(3\right)????$$

Commented by bobhans last updated on 28/May/20

$$x=e^{\ln \left(\mathrm{x}\right)}$$$$3^{-{4\mathrm{x}}^2}={\mathrm{e}}^{\ln {\left(3\right)}^{-{4\mathrm{x}}^2}}={\mathrm{e}}^{-{4\mathrm{x}}^2\ln \left(3\right)}$$

Commented by Abdulrahman last updated on 28/May/20

$$\mathrm{can}\mathrm{you}\mathrm{explian}\mathrm{this}$$$$\int {\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dx}=?$$

Commented by prakash jain last updated on 28/May/20

https://en.m.wikipedia.org/wiki/Gaussian_integral This explain gaussian integral

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