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Integrating e^(-x^2)

N

Narmer

Diamond Member
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.
 
sqrt(pi)/2*erf(x) ?

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

 
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?
 
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.
Click to expand...

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.
 
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.
Click to expand...

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

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.
Click to expand...
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.

 
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.
Click to expand...

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.
Click to expand...
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.
Click to expand...

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

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.
Click to expand...
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.
Click to expand...

Just as my previous post was invisible to you, apparently. 😛
Click to expand...
I'm blind.

 
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