Gaussian Integral

By Gavin Adams on Thursday, January 17, 2002 - 02:18 am:

A = \int_{- \infty}^{\infty} e^{- x^{2}} dx

is the well known gaussian integral.

If we say that $B = \int_{- \infty}^{\infty} e^{- y^{2}} dy$ then:

$A^{2} = A B = \int_{- \infty}^{\infty} e^{- x^{2}} dx \int_{- \infty}^{\infty} e^{- y^{2}} dy$

NOW

Here are the two points that I don't understand:

Why can we then say that:
$\int_{- \infty}^{\infty} e^{- x^{2}} dx \int_{- \infty}^{\infty} e^{- y^{2}} dy = \int_{- \infty}^{\infty} e^{- x^{2} - y^{2}} dx dy$
How can we make this statement?
Why can we move the inside of the integral in $y$ into the integral in $x$ ?
After we do this we convert to a polar equation:
$\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- x^{2} - y^{2}} dx dy = \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r dr d θ$
I see that obviously $r^{2} = x^{2} + y^{2}$ by why do we multiply by $r dr d θ$ ?
How does one convert an equation like this generally into polar form to find an integral in rectangular form?

After that it's pretty easy to evaluate the integral, but i really don't understand those two essential points

By Dan Goodman on Thursday, January 17, 2002 - 10:00 pm:

I'm in a bit of a hurry so this answer will probably be too short.

In answer to (1): since $\int_{- \infty}^{\infty} e^{- x^{2}} dx$ is just a number $A$ (or $B$ ) and we know that for any integral and constant $K$ we have that $K \int f (y) dy = \int K . f (y) dy$ then we can set $f (y) = e^{- y^{2}}$ and $K = A$ to get that $\int_{- \infty}^{\infty} e^{- x^{2}} dx \int_{- \infty}^{\infty} e^{- y^{2}} dy = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- x^{2}} dx e^{- y^{2}} dy$ . That's just moving a constant inside the integral. Now, as far as the $\int_{- \infty}^{\infty} e^{- x^{2}} dx$ is concerned, $e^{- y^{2}}$ is just a constant, so we $\int_{- \infty}^{\infty} e^{- x^{2}} dx e^{- y^{2}} = \int_{- \infty}^{\infty} e^{- x^{2}} e^{- y^{2}} dx = \int_{- \infty}^{\infty} e^{- x^{2} - y^{2}} dx$ . So we get that $\int_{- \infty}^{\infty} e^{- x^{2} - y^{2}} dx dy$ . So, there are two steps involved, each is just taking a constant inside an integral.

The answer to (2) is in general a bit more complicated. There was a discussion not that long ago on Nrich about Jacobians and change of coordinates in integrals (which is what you want to know about), perhaps someone will post the link.

The basic idea is this:

Suppose we had an integral with respect to $x$ . If we change variables to $y = 2 x$ then we have to change the limits first of all, but also change the $dx$ to $(1 / 2) dy$ . Intuitively, you can say that $dy / dx = 2$ so $dy = 2 dx$ so $dx = dy / 2$ . This is sort of dodgy but gives the right answer. What we're doing is scaling the area of the infinitessimal element by a certain factor, in this case 1/2. You have to include that scale factor when you change variables. It turns out that the right scale factor for the change from 2D cartesian (or rectangular if you prefer) to 2D polar coordinates is $r$ . If you draw a box with vertices $(r, θ)$ , $(r + dr, θ)$ , $(r, θ + d θ)$ and $(r + dr, θ + d θ)$ and work out its area, this will be the scaling factor at the point $(r, θ)$ . This is true for cartesian coordinates as well, but the box obviously has area $dx . dy$ in that case. Have a go at working it out, remember that you can discard any terms with more that three $d$ terms, so $dr . dr . d θ$ can be discarded for example, because when you integrate it over $r$ and $θ$ you have a $dr$ left over which makes it 0. Sorry I can't be more clear but I'm short of time...

By Brad Rodgers on Thursday, January 17, 2002 - 10:03 pm:

For 1), we know that A is a constant, so

A \int_{0}^{\infty} e^{- y^{2}} dy = \int_{0}^{\infty} A e^{- y^{2}} dy

In that way, it's pretty intuitional. There is a rigorous theorem about this called Fubini's theorem, but it involves special cases, so isn't neccessary here.

For 2), to understand it rigorously, you would have to understand change of variables in a double integral, which involves something called a Jacobian. I can try to find a link on this though, if you want. You can understand it pretty well though, by considering that the area of a region in rectangular coordinates is

\int \int_{R} dx dy

whereas the area in polar coordinates is

$\int_{a}^{b} (1 / 2) r^{2} d θ = \int \int_{R'} r dr d θ$

where $R'$ is the same basic regionas $R$ , just $R'$ has representation in polar coordinates, and $R$ has representation in rectangular (e.g. if $R$ is $x^{2} + y^{2} = 1$ , then $R'$ is $r = 1$ ). Does that make sense? If not, I can try to find another explanation on another site.

If anyone else wants to explain this in a slightly different way, feel free to do so, as I generally have trouble putting concepts like this into words...

Brad

By Brad Rodgers on Thursday, January 17, 2002 - 10:03 pm:

Oops, hadn't yet seen your post, Dan.

Brad

By Yatir Halevi on Saturday, January 19, 2002 - 12:12 pm:

How, from that do you prove that the integral (from 0 to

\infty

) is equal to

\sqrt{π} / 2

Yatir

By Dan Goodman on Saturday, January 19, 2002 - 01:56 pm:

Yatir, once you have the integral from

- \infty

\infty

being

\sqrt{π}

it's immediate that the integral from 0 to

\infty

\sqrt{π} / 2

, because the integrand (the bit inside the integral) is symmetric about 0,

e^{- x^{2}} = e^{- (- x)^{2}}

By Yatir Halevi on Saturday, January 19, 2002 - 02:04 pm:

This I know.
My question was how do I integrate in polar coordinates?

Yatir

p.s.
Is there a different way of finding the value of the integral without switching to polar coordinates?

By Kerwin Hui on Saturday, January 19, 2002 - 02:31 pm:

Yatir,

Yes there is a way without switching to polar coordinates (but it requires more work). Let

defn

and show that f(x)+g(x)=pi/4. Now let x tends to infinity. See this archive thread for more details.

Alternatively we could start by a binomial distribution and let n tends to infinity. You can get the result after some (nasty) work. I saw that in a book which takes 2 pages to get to the answer with this method.

Kerwin