Question

If ${\int }_0^{\infty }{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}$, then ${\int }_0^{\infty }{e}^{-a{x}^2}dx$ where $a>0$ 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 $\frac{1}{2}\sqrt{\frac{\pi }{a}}$

We know that ${\int }_a^{b}f\left(x\right)dx=\int f\left(t\right)dt$

So, ${\int }_0^{\infty }{e}^{-x^2}dx={\int }_0^{\infty }{e}^{-t^2}dt=\frac{\sqrt{\pi }}{2}----\left(1\right)$

now in ${\int }_0^{\infty }{e}^{-a{x}^2}dx$, let $\sqrt{a}x=t$

now, $\sqrt{a}dx=dt$ (differentiation wrt $x$ , we get)

$\therefore dx=\frac{dt}{\sqrt{a}}$

So, here as $x\to 0,t\to 0$

& as $x\to \infty ,t\to \infty$

So ${\int }_0^{\infty }{e}^{-a{x}^2}dx={\int }_0^{\infty }{e}^{-t^2}\frac{dt}{\sqrt{a}}=\frac{1}{\sqrt{a}}{\int }_0^{\infty }{e}^{-t^2}dt$

From $\left(1\right)$

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

${\int }_0^{\infty }{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{2}}$