Question

lf ${\int }_0^{\infty }{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}$, then ${\int }_0^{\infty }{e}^{-a{x}^2}dx,a>0$ is

• A. $\frac{\sqrt{\pi }}{2}$
• B. $\frac{\sqrt{\pi }}{2a}$
• C. $2\frac{\sqrt{\pi }}{a}$
• D. $\frac{1}{2}\sqrt{\frac{\pi }{a}}$

Answer 1

The correct option is D $\frac{1}{2}\sqrt{\frac{\pi }{a}}$

${\int }_0^{\infty }{e}^{{-ax}^2}dx={\int }_0^{\infty }{e}^{-(\sqrt{a}x{)}^2}dx$

Let, $\sqrt{a}.x=1$

$\sqrt{a}.dx=dtdx=\frac{dt}{\sqrt{a}}I={\int }_0^{\infty }{e}^{-t^2}.\frac{dt}{\sqrt{a}}=\frac{1}{\sqrt{a}}{\int }_0^{\infty }{e}^{-t^2}dt=\frac{1}{\sqrt{a}}\times \frac{\sqrt{\pi }}{2}=\frac{1}{2}\sqrt{\frac{\pi }{a}}$

Hence the correct answer is $\frac{1}{2}\sqrt{\frac{\pi }{a}}$