Integrating e^(-x^2)

[Narmer]

The boundary is from negative to positive infinity. I know the answer is the square root of pi. I also know that it's good to switch to polar coordinate. But I cannot seem to get the right function to plug in the (new) boundaries. I haven't done calculus since high school and cannot remember this.

 

[FleshLight]

[e^(-x^2)]/[-x^(-2)]

 

[Narmer]

Originally posted by: FleshLight
[e^(-x^2)]/[-x^(-2)]

No, that's not right. You have to change it to polar coordinates, like: x = rcosy

 

[TheoPetro]

man F polar cords...

 

[BrownTown]

sqrt(pi)/2*erf(x) ?

edit, and oh yeah screw polar coordinates, this is why god invented calculators, or at least lookup tables.

 

[Random Variable]

It doesn't have a closed form solution. Expand it in a Taylor series.

 

[TuxDave]

I don't think there is one. You have to do numerical analysis to get the approximation.

 

[mfs378]

This is how you do it:

The integral you want to solve is

int_{-inf}^{inf} exp(-x^2) dx

This is equal to

sqrt( (int_{-inf}^{inf} exp(-x^2) dx) * (int_{-inf}^{inf} exp(-y^2) dy))

Now convert into polar coordinates and it should be straightforward from there.

 

[Throwmeabone]

Yeah, I thought it doesn't have an antiderivative.

 

[mfs378]

Originally posted by: Throwmeabone
Yeah, I thought it doesn't have an antiderivative.

Yeah it doesn't but you can still integrate it if the limits are infinite.

 

[Random Variable]

I found this in some of my old notes:

let I = integral[e^(-z^2)]dz

then I^2 = integral[e^(-x^2)]dx*integral[(e^(-y^2)]dy

write the iterated integral as a double integral:

I^2 = integral[e^(-(x^2+y^2))]dxdy

change to polar coordinates:

I^2=integral[e^(-r^2)*r]drd?

then I^2 = Pi (using the proper limits)

and I=sqrt(Pi)

 

[Narmer]

Thanks guys, that solution has been staring me in the face for hours but I thought I couldn't do it that way.

EDIT: By the way are the proper limits for theta -pi to pi and -1 to 1 for the radius?

 

[chuckywang]

Originally posted by: Narmer
The boundary is from negative to positive infinity. I know the answer is the square root of pi. I also know that it's good to switch to polar coordinate. But I cannot seem to get the right function to plug in the (new) boundaries. I haven't done calculus since high school and cannot remember this.

Let I = integral e^(-x^2) dx from -inf to inf.

Therefore, I^2 = integral e^(-x^2) dx from -inf to inf * integral e^(-y^2) dy from -inf to inf

= double integral of e(-(x^2+y^2)) from -inf to -inf dx and -inf to inf dy

Converting to polar coordinates, we get: integral of r*e^(-r^2) from -inf to inf dr and 0 to 2pi d(theta). Integrating that we get that I^2 = pi, so I = sqrt (pi).

Hope that helps.

 

[sao123]

Originally posted by: Narmer
The boundary is from negative to positive infinity. I know the answer is the square root of pi. I also know that it's good to switch to polar coordinate. But I cannot seem to get the right function to plug in the (new) boundaries. I haven't done calculus since high school and cannot remember this.

The TI 89 says it cannot be integrated... who are you to question this wisdom?

 

[Random Variable]

Originally posted by: chuckywang

Originally posted by: Narmer
The boundary is from negative to positive infinity. I know the answer is the square root of pi. I also know that it's good to switch to polar coordinate. But I cannot seem to get the right function to plug in the (new) boundaries. I haven't done calculus since high school and cannot remember this.

Let I = integral e^(-x^2) dx from -inf to inf.

Therefore, I^2 = integral e^(-x^2) dx from -inf to inf * integral e^(-y^2) dy from -inf to inf

= double integral of e(-(x^2+y^2)) from -inf to -inf dx and -inf to inf dy

Converting to polar coordinates, we get: integral of r*e^(-r^2) from -inf to inf dr and 0 to 2pi d(theta). Integrating that we get that I^2 = pi, so I = sqrt (pi).

Hope that helps.

I guess my posts are invisible because I posted the exact same thing about an hour ago.

And of course r goes from 0 to infinity as r is never negative.

 

[mfs378]

Originally posted by: Random Variable

Originally posted by: chuckywang

Originally posted by: Narmer
The boundary is from negative to positive infinity. I know the answer is the square root of pi. I also know that it's good to switch to polar coordinate. But I cannot seem to get the right function to plug in the (new) boundaries. I haven't done calculus since high school and cannot remember this.

Let I = integral e^(-x^2) dx from -inf to inf.

Therefore, I^2 = integral e^(-x^2) dx from -inf to inf * integral e^(-y^2) dy from -inf to inf

= double integral of e(-(x^2+y^2)) from -inf to -inf dx and -inf to inf dy

Converting to polar coordinates, we get: integral of r*e^(-r^2) from -inf to inf dr and 0 to 2pi d(theta). Integrating that we get that I^2 = pi, so I = sqrt (pi).

Hope that helps.

I guess my posts are invisible because I posted the exact same thing about an hour ago.

And of course r goes from 0 to infinity as r is never negative.

Just as my previous post was invisible to you, apparently.

 

[BrownTown]

too bad you people are all wrong and I'm right

 

[Random Variable]

Originally posted by: BrownTown
too bad you people are all wrong and I'm right

All your answer says is that int[e^(-x^2)]dx = (sqrt(Pi)/2)*2/sqrt(Pi)*int[e^(-x^2)]dx.

 

[Random Variable]

Originally posted by: mfs378

Originally posted by: Random Variable

Originally posted by: chuckywang

Originally posted by: Narmer
The boundary is from negative to positive infinity. I know the answer is the square root of pi. I also know that it's good to switch to polar coordinate. But I cannot seem to get the right function to plug in the (new) boundaries. I haven't done calculus since high school and cannot remember this.

Let I = integral e^(-x^2) dx from -inf to inf.

Therefore, I^2 = integral e^(-x^2) dx from -inf to inf * integral e^(-y^2) dy from -inf to inf

= double integral of e(-(x^2+y^2)) from -inf to -inf dx and -inf to inf dy

Converting to polar coordinates, we get: integral of r*e^(-r^2) from -inf to inf dr and 0 to 2pi d(theta). Integrating that we get that I^2 = pi, so I = sqrt (pi).

Hope that helps.

I guess my posts are invisible because I posted the exact same thing about an hour ago.

And of course r goes from 0 to infinity as r is never negative.

Just as my previous post was invisible to you, apparently.

I'm blind.