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 $$\int_0^\infty e^{-a\log(2)x^2}\,dx=\sqrt{\frac{\pi}{a\log 2}}.$$
Starting from where you left off:
$$I=\int_{0}^{\infty}2^{-at}t^{1/2-1}dt$$
$$I=\int_{0}^{\infty}\exp(\ln(2^{-at}))t^{\frac12-1}dt=\int_{0}^{\infty} e^{-at\ln(2)}t^{\frac12-1}dt $$
while the gamma function may be defined as:
$$\Gamma(x)\mathop=^\text{def}\int_0^\infty t^{z-1} e^{-t} dt$$
Now try $$at\ln(2)=x\implies dt=\frac{dx}{a\ln(2)}$$
with the bounds remaining the same for defined $a\ne0$:
$$\int_{0}^{\infty} e^{-at\ln(2)}t^{\frac12-1}dt = \int_{0}^{\infty} e^{-x}\left(\frac x{a\ln(2)}\right)^{-\frac12}\frac {dx}{a\ln(2)}= \frac1{\sqrt{a\ln(2)}}\int_{0}^{\infty} e^{-x}x^{\frac12-1}dx=\sqrt{\frac\pi{a\ln(2)}}$$ with