Evaluate $\int_{0}^{\infty}2^{-ax^2}dx$ by using Gamma function


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?

  • 5
    $\begingroup$ Note that $$a^x=e^{\ln(a)x}$$ then use the gamma function definition $\endgroup$ Oct 29, 2021 at 14:34
  • $\begingroup$ What definition do you use for the "Gamma function"? $\endgroup$
    – user765539
    Oct 29, 2021 at 14:38
  • 1
    $\begingroup$ @user:britannica.com/science/gamma-function $\endgroup$
    – P.Styles
    Oct 29, 2021 at 14:39
  • 2
    $\begingroup$ @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. $\endgroup$
    – user765539
    Oct 29, 2021 at 14:41
  • $\begingroup$ @user:$\sqrt \pi /(ln 2^a)^{1/2}$?? $\endgroup$
    – P.Styles
    Oct 29, 2021 at 14:47

2 Answers 2


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

  • 1
    $\begingroup$ This is the error function (+1) $\endgroup$ Oct 29, 2021 at 15:08
  • 2
    $\begingroup$ I would rather call that the Gauss integral! $\endgroup$
    – LL 3.14
    Oct 29, 2021 at 15:10

Starting from where you left off:


$$I=\int_{0}^{\infty}\exp(\ln(2^{-at}))t^{\frac12-1}dt=\int_{0}^{\infty} e^{-at\ln(2)}t^{\frac12-1}dt $$

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

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


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