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Question Number 43147 by rahul 19 last updated on 07/Sep/18

$$\mathrm{If}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-x^2}dx=\frac{\sqrt{\pi }}{2},$$$$\mathrm{then}provethat{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{a}{x}^2}dx=\sqrt{\frac{\pi }{4\mathrm{a}}}$$$$\mathrm{where}\mathrm{a}>0.$$

Commented by MrW3 last updated on 07/Sep/18

$$lett=\sqrt{a}x$$$$dt=\sqrt{a}dx$$$$dx=\frac{dt}{\sqrt{a}}$$$${\int }_0^{\mathrm{\infty }}{e}^{-{ax}^2}dx$$$$={\int }_0^{\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{a}}$$$$=\frac{1}{\sqrt{a}}{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt$$$$=\frac{1}{\sqrt{a}}\times \frac{\sqrt{\pi }}{2}$$$$=\sqrt{\frac{\pi }{4a}}$$

Commented by rahul 19 last updated on 07/Sep/18

thank you sir ��

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