Evaluate $\int_{0}^{\infty}2^{-ax^2}dx$ by using Gamma function Evaluate  $\int_{0}^{\infty}2^{-ax^2}dx$ by using Gamma function
$$I=\int_{0}^{\infty}2^{-ax^2}dx$$
Solution:$$ \text{Let} \\x^2=t\implies 2xdx=dt\implies dx=\frac{dt}{2x}\implies dx=\frac{dt}{2\sqrt t}$$
$$I=\int_{0}^{\infty}2^{-at}\frac{1}{2\sqrt t}dt$$
$$\implies I=\int_{0}^{\infty}2^{-at}t^{1/2-1}dt$$
what should be the next step?

Your integral is just
$$\int_0^\infty e^{-a\log(2)x^2}\,dx=\sqrt{\frac{\pi}{a\log 2}}.$$
