Question

The value of ${\int }_0^{\infty }{\int }_0^{\infty }{e}^{-x^2}.e^{-y^2}dxdy$ is

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

Answer 1

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

Given integral is
$I={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-x^2}.e^{-y^2}dxdy$
$={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-(x^2+y^2})dxdy$
Put $x=rcos\theta ,y=rsin\theta ,|J|=r$ (Jacobian of polar coordinate system).
Then to cover whole $1^{st}$ Quadrant $\theta$ varies from $0$ to $\frac{\pi }{2}$ and $r$ varies from $0$ to $\infty$
So, $I={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-(x^2+y^2)}.dxdy$
$={\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^{\infty }{e}^{-r^2}.\left(rdrd\theta \right)$
$=\frac{\pi }{2}{\int }_{r=0}^{\pi }r.e^{-r^2}.dr=\frac{\pi }{4}$ (Put $r^2=t$)