1
$\begingroup$

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?

$\endgroup$
6
  • 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

7
$\begingroup$

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

$\endgroup$
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
3
$\begingroup$

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

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.