Question Number 43147 by rahul 19 last updated on 07/Sep/18
$$\mathrm{If}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}\:}\:, \\ $$$$\mathrm{then}\:{prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{a}{x}^{\mathrm{2}} } {dx}\:=\:\sqrt{\frac{\pi}{\mathrm{4a}}} \\ $$$$\mathrm{where}\:\mathrm{a}>\mathrm{0}. \\ $$
Commented by MrW3 last updated on 07/Sep/18
$${let}\:{t}=\sqrt{{a}}{x} \\ $$$${dt}=\sqrt{{a}}{dx} \\ $$$${dx}=\frac{{dt}}{\:\sqrt{{a}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}^{\mathrm{2}} } {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } \frac{{dt}}{\:\sqrt{{a}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}}}×\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$=\sqrt{\frac{\pi}{\mathrm{4}{a}}} \\ $$
Commented by rahul 19 last updated on 07/Sep/18
thank you sir ��