Integrate e^-x^2

e^-x^2 is not integrable directly, so a transformation has to first be made. Assume J = int(e^-x^2 dx), and x=y. Therefore:

J^2 = int(e^-x^2 dx * e^-y^2 dy)
J^2 = int(e^-(x^2+y^2) dx dy)

Transform:

J^2 = int(e^-r^2 rdr) * int(d(theta))

Usually, at this point, various sources integrate it from -inf to inf, and obtain sqrt(pi) as the solution of int(e^-x^2 dx) [-inf,inf].

My question is, why should this only work unless the function is integrated from -inf to inf? Why can't it work for any value of x or r?

My question is, why should this only work unless the function is integrated from -inf to inf? Why can't it work for any value of x or r?

It does work for any value of x and r.

For instance, from x=0 to infinity would give you sqrt(Pi)/2. Or from 0 to 10 is .8862269255.

But other than very specific values, its hard to define the solution to the integral

What if I wanted to evaluate the integral from 0 to 1?

e^-x^2 is not integrable directly, so a transformation has to first be made. Assume J = int(e^-x^2 dx), and x=y. Therefore:

J^2 = int(e^-x^2 dx * e^-y^2 dy)
J^2 = int(e^-(x^2+y^2) dx dy)

Transform:

J^2 = int(e^-r^2 rdr) * int(d(theta))

Usually, at this point, various sources integrate it from -inf to inf, and obtain sqrt(pi) as the solution of int(e^-x^2 dx) [-inf,inf].

My question is, why should this only work unless the function is integrated from -inf to inf? Why can't it work for any value of x or r?

Because you still have to evaluate this function (Note that you still have the in there, which you in the beginning wanted to transform to some other expression):


As r goes toward infinity and negative infinity, goes toward zero. This isn't true for real numbers, and thus we have an issue.

What if I wanted to evaluate the integral from 0 to 1?

Use the error function, or extrapolate it yourself. I recommend the error function.

Noah

As r goes toward infinity and negative infinity, goes toward zero. This isn't true for real numbers, and thus we have an issue.

What? What do you mean it's not true? I don't understand what you just did here.

What? What do you mean it's not true? I don't understand what you just did here.

I meant that

and that


Noah

The problem with evaluating that function on a region different than the whole (x,y) plane is that it will generally lead to something difficult to describe in polar coordinates (r,theta) which in turn lead to an integral that you cannot separate in two independent ones (one in r, one in theta).

The way you do it, you pass from a section (or the whole) x axis to a symetrical thing in 2D, which will most likely be in the form of a rectangle, which is hard to describe in polar.

+C

Hahahawhat

Kind of unrelated to Harrisonized's problem, but still in Calculus.

Integral (x / (4 + ln x)) from 4 to x. (Quite frankly, the problem is find G'(e) if G(x) = Integral (e ^( ln x) / (4 + ln x)) from 4 to x, but the first step is to find the integration of that, and the e ^ ln x cancels out to x, right?)

I plugged it into Wolfram Alpha and got a crazy answer. Unless that's the answer...I'm screwed.

Use your own damn thread, Corn.

I meant that

and that

I still don't understand this. e^-x^2 only ever hits 0 at inf, so I don't know what you're trying to point out.

The way you do it, you pass from a section (or the whole) x axis to a symetrical thing in 2D, which will most likely be in the form of a rectangle, which is hard to describe in polar.

Okay, this is more helpful. But it makes me wonder. When the function is transformed to polar coordinates, it becomes this:



r is just the function's position from zero. The volume of a cylinder is 2pi r h. Therefore, I could find the volume under the curve from r1 to r2, and since the volume of a cylinder is 2pi r h, I could divide to get the area underneath the curve?

I'm talking about applying this. How would I apply this to get the area under e^-x^2?

I still don't understand this. e^-x^2 only ever hits 0 at inf, so I don't know what you're trying to point out.

Look here, this shows the proof - and why I stated that infinity and negative infinity is important end-points for the function you're integrating.
http://en.wikipedia.org/wiki/Gaussian_integral

Noah

after you have that square region of [-1;1] for both x and y you have two integrals on x and y with limits -1 and 1. To transform that into the polar integral you have to change the limits of the integrals into something that describes that square region. You'd probably have to break the integral and in 4 parts, theta belonging to [-pi/4;pi/4], [pi/4;3pi/4], [3pi/4;5pi/4], [5pi/4;-pi/4] and have r go from zero until the the functions x=1, y=1, x=-1 and y=-1 respectively (and change those to polar coordinates).

So the limits of your integral on r would depend on theta so you wouldn't be able to separate the two integrals and you have a much more complex problem.

Okay, this is more helpful. But it makes me wonder. When the function is transformed to polar coordinates, it becomes this:



r is just the function's position from zero. The volume of a cylinder is 2pi r h. Therefore, I could find the volume under the curve from r1 to r2, and since the volume of a cylinder is 2pi r h, I could divide to get the area underneath the curve?

I'm talking about applying this. How would I apply this to get the area under e^-x^2?

if you want the volume of a ring e^-r^2, for example 1 <= r <= 2, do 2 integrals. take the limits of one from 0 <= r <= 1 and the other as 0 <= r <= 2. subtract the result and integrate over theta (the r integrals give you the area under a slice of the sheet) to find the volume of the ring.

...im going to assume that you arent trying to find a volume with nonseparable limits like that nice mathematica picture you have there.

It's been a couple (4) years since I've taken calc, but couldn't you take ln(e^-x^2) = -x^2?

Or is that the derivative?

It's been a couple (4) years since I've taken calc, but couldn't you take ln(e^-x^2) = -x^2?

Or is that the derivative?

I think you're thinking of how you derivate 2^x (or some other constant instead of 2), you take exp(ln(2^x)) because exp(ln( f(x))) = f(x). Normally, you can't just take natural logs without an taking an exponential too because then you have a different function

Forget what I said it's wrong.

^ Even if it does, none of that helps you integrate the original function. As I said before, I'm fairly sure you can't just go around making changes to the function that aren't equal to 1, and expect the same answer. (Its why when you're doing some integrals, you take natural logs AND exponentials at the same time, because log(exp f(x))= f(x), or when you're working with complex numbers, you multiply by the complex conjugate/ complex conjugate, because its equal to 1.) if you take a natural log, without taking an exponential, you've changed the function, so the integral won't give you the correct answer for the function anymore.

E.g. If I just take the logarithm of some function, for simplicity, lets say y=2x
y=2x
ln(y)=ln(2x)
int(ln(y))=int(ln(2x))
yln(y) - y = ln(2x)*x -x +C

Which clearly is not equal to int (y) = int(2x) [ y^2/2 = x^2+C], because, for one thing, the natural logarithms are undefined for x or y less than 0, where clearly the original function doesn't have that problem

A transformation of coordinates system does not change the integral, provided the new system is compatible with the old system. Can be shown through PDEs, and the transformation is easily done with Jacobian. We just make our life easier through transformation since we can't exactly do the integral in the cartesian system.

I'm not a mathmo so I can't handle too much of this.

Hadriel