Answer
a) $\dfrac{\sqrt{\pi}}{2}$ b) $1$
Work Step by Step
Our aim is to integrate the integral as follows:
A) $ I^2 =\int^{\infty}_0\int^{\infty}_0 e^{-(x^2+y^2)} \space dx \space dy =\int^{\pi/2}_ 0 \int^{\infty}_0(e^{-r^2})r \space dr \space d\theta =\int^{\pi/2}_0 [\lim\limits_{b \to \infty}\int^b_ re^{-r^2}] \space d\theta =-\dfrac{1}{2} \int^{\pi/2}_0 \lim\limits_{b \to \infty}(e^{-b^2}-1) \space d\theta =\dfrac{1}{2} \times \int^{\pi/2}_0 d\theta =\dfrac{\pi}{4} =\dfrac{\sqrt{\pi}}{2}$
B) Now, $\lim\limits_{x \to \infty} \int^x_0 \dfrac{2e^{-t^2}}{\sqrt{\pi}} \space dt =\dfrac{2}{\sqrt{\pi}} \times \int^\infty_0 e^{\it^2} dt \space =[\dfrac{2}{\sqrt{\pi}}(\dfrac{\sqrt{\pi}}{2})]=1$
