# 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?

• Note that $$a^x=e^{\ln(a)x}$$ then use the gamma function definition Oct 29, 2021 at 14:34
• What definition do you use for the "Gamma function"?
– user765539
Oct 29, 2021 at 14:38
• Oct 29, 2021 at 14:39
• @P.Styles: For the next step, can you compare the $I$ you got to the form of the Gamma function? The answer is almost there.
– user765539
Oct 29, 2021 at 14:41
• @user:$\sqrt \pi /(ln 2^a)^{1/2}$?? Oct 29, 2021 at 14:47

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

• This is the error function (+1) Oct 29, 2021 at 15:08
• I would rather call that the Gauss integral! Oct 29, 2021 at 15:10

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

$$\Gamma\left(\frac12\right)=\sqrt \pi$$