Question

The integral $\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{e}^{-x^2/2}dx$ is equal to

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

Answer 1

The correct option is C $1$

$I=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{e}^{-x^2/2}dx$
$=\frac{2}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{-x^2/2}.dx$
Put $\frac{x^2}{2}=t$
$I=\frac{2}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{-t}.\frac{dt}{\sqrt{2}\sqrt{t}}$
$=\frac{1}{\sqrt{2\pi }}\times 2{\int }_0^{\infty }{e}^{-t}\frac{1}{\sqrt{2}}{t}^{-1/2}dt$