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Gaussian integral

  • #1
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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
 

Answers and Replies

  • #2
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I figured it out

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

still a bit puzzled on how I show that though
 
  • #3
SteamKing
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Have you tried substitution?
 
  • #4
exo
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You can try setting it up as a double integral and doing it in polar coordinates.
 
  • #5
LCKurtz
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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???
 
  • #6
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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"
 
  • #7
exo
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Guess I am going to give it a go.You can set it up as

[itex]\displaystyle\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-8x^{2}-8y^{2}}dxdy[/itex]

Now this can be rewritten as

[itex]\displaystyle\int^{\infty}_{-\infty}e^{-8y^{2}}\int^{\infty}_{-\infty}e^{-8x^{2}}dxdy[/itex]

If we label the original integral [itex]A[/itex] solving this double integral will give us [itex]A^{2}[/itex], 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

[itex]\displaystyle\int^{\infty}_{0}\int^{2\pi}_{0}re^{-8r^{2}}d\theta dr = [/itex]

[itex]= 2 \pi \displaystyle\int^{\infty}_{0}re^{-8r^{2}}dr = [/itex]

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

So [itex]A = \sqrt{\frac{\pi}{8}}[/itex]
 
Last edited:
  • #8
HallsofIvy
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No, using the fact that [itex]\int 3 f(x)dx= 3\int f(x)dx[/itex] is NOT cheating!

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

Do you know how to show that [itex]\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi}[/itex]? If not, "set it up as a double integral and change to polar coordinates" as exo suggested. Once you know that, because you have [itex]e^{-ax^2}[/itex] and want [itex]e^{-u^2}[/itex] use the substitution [itex]-u^2= ax^2[/itex] or [itex]u= \sqrt{a} x[/itex].
 
  • #9
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No, using the fact that [itex]\int 3 f(x)dx= 3\int f(x)dx[/itex] is NOT cheating!

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

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

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.
 
  • #10
SammyS
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...

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

I'm intrigued .
 
  • #11
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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##
 
  • #12
HallsofIvy
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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 "[itex]\lim_{a\to\infty}\int_0^a[/itex]".

##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##
 
  • #13
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Oops, yeah thanks. It was the end of the night. Its worked out properly on paper.
 
Last edited:

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