What is the solution to the Gaussian integral?

[jonroberts74]

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

 

[jonroberts74]

I figured it out

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

still a bit puzzled on how I show that though

 

[SteamKing]

Have you tried substitution?

 

[exo]

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

 

[LCKurtz]

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?

 

[jonroberts74]

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}}

 

[HallsofIvy]

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.

 

[jonroberts74]

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, more so cheating myself of the fun of solving a neat integral.

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

 

[SammyS]

...

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

Let's see the details of that!

I'm intrigued .

 

[jonroberts74]

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]

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".

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

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

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

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

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

 

[jonroberts74]

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