# Gaussian integral

## Homework Statement

I am asked to evaluate ##\displaystyle\int_{-\infty}^{\infty} 3e^{-8x^2}dx##

## Homework Equations

I know

##\displaystyle\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}##

## The Attempt at a Solution

based on an example in the book it seems a change of variables is what I need to do.

I'm just not sure how

Related Calculus and Beyond Homework Help News on Phys.org
I figured it out

##3\sqrt{\frac{\pi}{8}}##

still a bit puzzled on how I show that though

SteamKing
Staff Emeritus
Homework Helper
Have you tried substitution?

exo
You can try setting it up as a double integral and doing it in polar coordinates.

LCKurtz
Homework Helper
Gold Member
I figured it out

##3\sqrt{\frac{\pi}{8}}##

still a bit puzzled on how I show that though
Now I'm puzzled. You figured it out but now you don't know how you did it???

Now I'm puzzled. You figured it out but now you don't know how you did it???
I used what I knew of the gaussian integral

and when it is of the form ##\displaystyle\int_{-\infty}^{\infty} e^{-ax^2}dx##

the answer is ##\sqrt{\frac{\pi}{a}}## and the 3 is just a constant so that factored out of the integral

so ##3\sqrt{\frac{\pi}{8}}##

I'll try it with polar coordinates though, this way was a bit like "cheating"

exo
Guess I am going to give it a go.You can set it up as

$\displaystyle\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-8x^{2}-8y^{2}}dxdy$

Now this can be rewritten as

$\displaystyle\int^{\infty}_{-\infty}e^{-8y^{2}}\int^{\infty}_{-\infty}e^{-8x^{2}}dxdy$

If we label the original integral $A$ solving this double integral will give us $A^{2}$, so if we manage to solve it we can just take the square root and get our answer.

Let's use polar coordinates then our integral becomes

$\displaystyle\int^{\infty}_{0}\int^{2\pi}_{0}re^{-8r^{2}}d\theta dr =$

$= 2 \pi \displaystyle\int^{\infty}_{0}re^{-8r^{2}}dr =$

$= \left[ 2 \pi \frac{(-e^{-8r^{2}})}{16} \right]_{0}^{\infty} = \frac{\pi}{8}$

So $A = \sqrt{\frac{\pi}{8}}$

Last edited:
HallsofIvy
Homework Helper
No, using the fact that $\int 3 f(x)dx= 3\int f(x)dx$ is NOT cheating!

If you are concerned about how they got that $\int_{-\infty}^\infty e^{-ax^2} dx= \sqrt{\pi/a}$-

Do you know how to show that $\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi}$? If not, "set it up as a double integral and change to polar coordinates" as exo suggested. Once you know that, because you have $e^{-ax^2}$ and want $e^{-u^2}$ use the substitution $-u^2= ax^2$ or $u= \sqrt{a} x$.

No, using the fact that $\int 3 f(x)dx= 3\int f(x)dx$ is NOT cheating!

If you are concerned about how they got that $\int_{-\infty}^\infty e^{-ax^2} dx= \sqrt{\pi/a}$-

Do you know how to show that $\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi}$? If not, "set it up as a double integral and change to polar coordinates" as exo suggested. Once you know that, because you have $e^{-ax^2}$ and want $e^{-u^2}$ use the substitution $-u^2= ax^2$ or $u= \sqrt{a} x$.

I didn't mean cheating was about pulling the constant outside the integral. I meant I'd rather work the problem out properly than use a known formula to get skip to the end. And of course that isn't cheating either, moreso cheating myself of the fun of solving a neat integral.

I did it using polar coordinates, I got to the same answer.

SammyS
Staff Emeritus
Homework Helper
Gold Member
...

I did it using polar coordinates, I got to the same answer.
Let's see the details of that!

I'm intrigued .

Let's see the details of that!

I'm intrigued .

I'll let G = ##\int_{-\infty}^{\infty} 3e^{-8x^2}dx##

and ##G^2 =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}9e^{-8(x^2+y^2)}dxdy##

Changing to polar

##\int_{0}^{2\pi} \int_{0}^{\infty}9e^{-8r^2}rdrd\theta##

the outside integral doesn't rely and is not affect by the one on the inside

##\int_{0}^{2\pi} d\theta \int_{0}^{\infty}9e^{-8r^2}rdr##

I'm going to introduce the limit now, previously left out for expedience

##2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}9e^{-8r^2}rdr\Bigg]##

##9*2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}e^{-8r^2}rdr\Bigg]##

##u=8r^2; du=16rdr \rightarrow \frac{du}{16r}=dr##

##9*2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}e^{-u^2}r\frac{du}{16r}\Bigg]##

##9*2\pi\Bigg[\lim_{a \to \infty}\frac{1}{16}\int_{0}^{\infty}e^{-u^2}du\Bigg]##

##9*2\pi \Bigg[\lim_{a \to \infty}\frac{1}{16}(-e^{-8x^2})\Bigg|_{0}^{a}\Bigg]##

##9\Bigg[\lim_{a \to \infty}\frac{\pi}{8}(-e^{-8a^2}+e^{0})\Bigg]##

##9\cdot\frac{\pi}{8} = G^2##

##3\sqrt{\frac{\pi}{8}}=G##

HallsofIvy
Homework Helper
I'll let G = ##\int_{-\infty}^{\infty} 3e^{-8x^2}dx##

and ##G^2 =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}9e^{-8(x^2+y^2)}dxdy##

Changing to polar

##\int_{0}^{2\pi} \int_{0}^{\infty}9e^{-8r^2}rdrd\theta##

the outside integral doesn't rely and is not affect by the one on the inside

##\int_{0}^{2\pi} d\theta \int_{0}^{\infty}9e^{-8r^2}rdr##

I'm going to introduce the limit now, previously left out for expedience

##2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}9e^{-8r^2}rdr\Bigg]##

##9*2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}e^{-8r^2}rdr\Bigg]##

##u=8r^2; du=16rdr \rightarrow \frac{du}{16r}=dr##

##9*2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}e^{-u^2}r\frac{du}{16r}\Bigg]##
You mean "$\lim_{a\to\infty}\int_0^a$".

##9*2\pi\Bigg[\lim_{a \to \infty}\frac{1}{16}\int_{0}^{\infty}e^{-u^2}du\Bigg]##

##9*2\pi \Bigg[\lim_{a \to \infty}\frac{1}{16}(-e^{-8x^2})\Bigg|_{0}^{a}\Bigg]##

##9\Bigg[\lim_{a \to \infty}\frac{\pi}{8}(-e^{-8a^2}+e^{0})\Bigg]##

##9\cdot\frac{\pi}{8} = G^2##

##3\sqrt{\frac{\pi}{8}}=G##

Oops, yeah thanks. It was the end of the night. Its worked out properly on paper.

Last edited: