# Gaussian Integral

• Apr 15th 2010, 03:44 PM
Anthonny
Gaussian Integral
Hi
I am trying to learn the proof for the value of the Gaussian Integral posted here: Gaussian integral - Wikipedia, the free encyclopedia

under 'By Cartesian Coordinates"

I understand the basics of double integration, but I don't quite understand this proof.

I see that the function is an even function...but what is the significance of saying "Let x=ys and dy=xds"

I also cannot see how
$4\int^\infty_0\int^\infty_0e^{-(x^2+y^2)}dydx$
arises when
$2\int^\infty_0e^{-x^2}dx$ is squared.

And...I also do not understand how the 's' variable arises later on into the proof.

I would greatly appreciate if these questions were answered and it would be extremely appreciated if a quick run through of this proof was posted.

Thank you very much.
• Apr 15th 2010, 04:15 PM
AllanCuz
Quote:

Originally Posted by Anthonny
Hi
I am trying to learn the proof for the value of the Gaussian Integral posted here: Gaussian integral - Wikipedia, the free encyclopedia

under 'By Cartesian Coordinates"

I understand the basics of double integration, but I don't quite understand this proof.

I see that the function is an even function...but what is the significance of saying "Let x=ys and dy=xds"

I also cannot see how
$4\int^\infty_0\int^\infty_0e^{-(x^2+y^2)}dydx$
arises when
$2\int^\infty_0e^{-x^2}dx$ is squared.

And...I also do not understand how the 's' variable arises later on into the proof.

I would greatly appreciate if these questions were answered and it would be extremely appreciated if a quick run through of this proof was posted.

Thank you very much.

Let

$I = 2\int^\infty_0e^{-x^2}dx$

$I^2 = (2\int^\infty_0e^{-x^2}dx)^2$

Of course we can say that y=x and this does not change anything. Therefore,

$2\int^\infty_0e^{-x^2}dx = 2\int^\infty_0e^{-y^2}dy$

Then,

$I^2 = (2\int^\infty_0e^{-x^2}dx)^2 = (2\int^\infty_0e^{-x^2}dx) (2\int^\infty_0e^{-y^2}dy) = 4\int^\infty_0\int^\infty_0e^{-(x^2+y^2)}dydx$

Of course we can then let

$r^2=x^2+y^2$

to complete this proof in cylindrical co-ordinates.

$4\int^\infty_0\int^\infty_0e^{-(x^2+y^2)}dydx = \int^\infty_{- \infty}\int^\infty_{- \infty} e^{-(x^2+y^2)}dydx =\iint re^{-(r^2)}drd \theta$

$\iint re^{-(r^2)}drd \theta = \int_0^{ 2\pi} d \theta \int_0^{ \infty } e^{-r^2}rdr$

Therefore,

$I^2 = 2 \pi \lim_{R \to +\infty} \frac{-1}{2} e^{-r^2}$

Of course this is evaluated from 0-->R

This then yields pi and we have

$I= \sqrt{\pi}$
• Apr 15th 2010, 06:15 PM
Anthonny
Quote:

Originally Posted by AllanCuz
Let

$I = 2\int^\infty_0e^{-x^2}dx$

$I^2 = (2\int^\infty_0e^{-x^2}dx)^2$

Of course we can say that y=x and this does not change anything. Therefore,

$2\int^\infty_0e^{-x^2}dx = 2\int^\infty_0e^{-y^2}dy$

Then,

$I^2 = (2\int^\infty_0e^{-x^2}dx)^2 = (2\int^\infty_0e^{-x^2}dx) (2\int^\infty_0e^{-y^2}dy) = 4\int^\infty_0\int^\infty_0e^{-(x^2+y^2)}dydx$

Of course we can then let

$r^2=x^2+y^2$

to complete this proof in cylindrical co-ordinates.

$4\int^\infty_0\int^\infty_0e^{-(x^2+y^2)}dydx = \int^\infty_{- \infty}\int^\infty_{- \infty} e^{-(x^2+y^2)}dydx =\iint re^{-(r^2)}drd \theta$

$\iint re^{-(r^2)}drd \theta = \int_0^{ 2\pi} d \theta \int_0^{ \infty } e^{-r^2}rdr$

Therefore,

$I^2 = 2 \pi \lim_{R \to +\infty} \frac{-1}{2} e^{-r^2}$

Of course this is evaluated from 0-->R

This then yields pi and we have

$I= \sqrt{\pi}$

Thank you!
I appreciate the thorough proof. However, this is the proof that involves polar coordination. I initially wanted an explanation for the alternative proof that involves what is posted here:

Gaussian integral - Wikipedia, the free encyclopedia

Under 'By Cartesian Coordinates"

And my main question here for this specific proof is what the significance of letting
"y=xs
dy=xds" has.

Along with this, I don't quite understand how the 's' arises in the proof.

However, I extremely appreciate your efforts in providing the proof for the Gaussian integral.
• Apr 15th 2010, 06:28 PM
AllanCuz
Quote:

Originally Posted by Anthonny
Thank you!
I appreciate the thorough proof. However, this is the proof that involves polar coordination. I initially wanted an explanation for the alternative proof that involves what is posted here:

Gaussian integral - Wikipedia, the free encyclopedia

Under 'By Cartesian Coordinates"

And my main question here for this specific proof is what the significance of letting
"y=xs
dy=xds" has.

Along with this, I don't quite understand how the 's' arises in the proof.

However, I extremely appreciate your efforts in providing the proof for the Gaussian integral.

This is no different then when I let y=x. As long as we clearly define our change of variables, and change our bounds accordingly, the integral/limit is still valid.

So, in this case

$y=xs$
$dy=xds$

were selected out of convenience. It then becomes clear that

$x^2+y^2 = x^2 + x^2s^2 = x^2(1+s)$

Which becomes our exponent. Of course you can then integrate the function quite easily because we have an x out front. They compute the integral directly, but you could most definately do By Parts here and get the same result.

Once this simplifies it comes to an integral of known form and is evaluated at that point.
• Apr 16th 2010, 02:31 PM
Anthonny
Quote:

Originally Posted by AllanCuz
This is no different then when I let y=x. As long as we clearly define our change of variables, and change our bounds accordingly, the integral/limit is still valid.

So, in this case

$y=xs$
$dy=xds$

were selected out of convenience. It then becomes clear that

$x^2+y^2 = x^2 + x^2s^2 = x^2(1+s)$

Which becomes our exponent. Of course you can then integrate the function quite easily because we have an x out front. They compute the integral directly, but you could most definately do By Parts here and get the same result.

Once this simplifies it comes to an integral of known form and is evaluated at that point.

I am starting to understand this proof. I can see why these variables are set and how they are replaced later on.

I just have one last small question.
When given:

$y=xs$
How does this turn into $dy=xds$?
Wouldn't it turn into $dy=dxds$?

I see how it is necessary for $dy=xds$ when it's substituted later on, but...I don't quite see why x remains unchanged.

Again, thank you very much. From your previous post I pretty much understand the proof.
• Apr 16th 2010, 02:35 PM
AllanCuz
Quote:

Originally Posted by Anthonny
I am starting to understand this proof. I can see why these variables are set and how they are replaced later on.

I just have one last small question.
When given:

$y=xs$
How does this turn into $dy=xds$?
Wouldn't it turn into $dy=dxds$?

I see how it is necessary for $dy=xds$ when it's substituted later on, but...I don't quite see why x remains unchanged.

Again, thank you very much. From your previous post I pretty much understand the proof.

Note that x does not depend on s and is therefore treated like a constant when we integrate our function with respect to s. So we then have

$\frac{dy}{ds} = (xs)^\prime$

With x being a constant and the derivative of s being one we have

$\frac{dy}{ds} = x$

Finally

$dy = xds$

I would like to note that you thought it might equal

$dy=dxds$

This assumes that x has a dependence on S, which it does not (we've only defined y in terms of s). But supposing that it did, say X WAS dependent on S. The above would still not be correct, we would need to use the product rule here

$\frac{dy}{ds} = xs + xs$

Clearly this is

$dy = (x+s)ds$

Which is not eqaul to what you had.
• Apr 16th 2010, 04:36 PM
Anthonny
Quote:

Originally Posted by AllanCuz
Note that x does not depend on s and is therefore treated like a constant when we integrate our function with respect to s. So we then have

$\frac{dy}{ds} = (xs)^\prime$

With x being a constant and the derivative of s being one we have

$\frac{dy}{ds} = x$

Finally

$dy = xds$

I would like to note that you thought it might equal

$dy=dxds$

This assumes that x has a dependence on S, which it does not (we've only defined y in terms of s). But supposing that it did, say X WAS dependent on S. The above would still not be correct, we would need to use the product rule here

$\frac{dy}{ds} = xs + xs$

Clearly this is

$dy = (x+s)ds$

Which is not eqaul to what you had.

Thanks! I now completely understand that y is dependent on s, and so x is treated like a constant.

I actually forgot to ask this question in my previous post, and I think this is my very last question, anyway the question is:

So we have that
$I^2=4\int^\infty_0{e^{-x^2}}dx\int^\infty_0{e^{-y^2}}dy$
is equal to:

$4\int^\infty_0\int^\infty_0{e^{-(x^2+y^2)}}dxdy$

and I understand why this works since $x=y$

But it states $y=xs$
So wouldn't $x=\frac{y}{s}$?

Thank you very much with the details on the proof. I'm really understanding this proof, I just need this one last clarification.
• Apr 16th 2010, 05:20 PM
AllanCuz
Quote:

Originally Posted by Anthonny
Thanks! I now completely understand that y is dependent on s, and so x is treated like a constant.

I actually forgot to ask this question in my previous post, and I think this is my very last question, anyway the question is:

So we have that
$I^2=4\int^\infty_0{e^{-x^2}}dx\int^\infty_0{e^{-y^2}}dy$
is equal to:

$4\int^\infty_0\int^\infty_0{e^{-(x^2+y^2)}}dxdy$

and I understand why this works since $x=y$

But it states $y=xs$
So wouldn't $x=\frac{y}{s}$?

Thank you very much with the details on the proof. I'm really understanding this proof, I just need this one last clarification.

It does, but why would we do this? It defeats the purpose of the subsitution in the first place.
• Apr 16th 2010, 06:52 PM
Anthonny
Quote:

Originally Posted by AllanCuz
It does, but why would we do this? It defeats the purpose of the subsitution in the first place.

So in this proof, it is true that
$x=y$
as well as
$y=xs$?